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Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression

November 10, 2023May 26, 2022 by Veerendra

Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 10 Arithmetic Progression

Arithmetic Progression Exercise 10A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
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Solution:

Question 2.
The nth term of sequence is (2n – 3), find its fifteenth term.
Solution:
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Question 3.
If the pth term of an A.P. is (2p + 3), find the A.P.
Solution:
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Question 4.
Find the 24th term of the sequence:
12, 10, 8, 6,……
Solution:
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Question 5.
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Solution:

Question 6.
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Solution:

Question 7.
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Solution:

Question 8.
Is 402 a term of the sequence :
8, 13, 18, 23,………….?
Solution:
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Question 9.
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Solution:

Question 10.
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Solution:

Question 11.
Which term of the A.P. 1 + 4 + 7 + 10 + ………. is 52?
Solution:
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Question 12.
If 5th and 6th terms of an A.P are respectively 6 and 5. Find the 11th term of the A.P
Solution:
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Question 13.

Solution:
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Question 14.
Find the 10th term from the end of the A.P. 4, 9, 14,…….., 254
Solution:
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Question 15.
Determine the arithmetic progression whose 3rd term is 5 and 7th term is 9.
Solution:
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Question 16.
Find the 31st term of an A.P whose 10th term is 38 and 10th term is 74.
Solution:
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Question 17.
Which term of the services :
21, 18, 15, …………. is – 81?
Can any term of this series be zero? If yes find the number of term.
Solution:
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Question 18.
An A.P. consists of 60 terms. If the first and the last terms be 7 and 125 respectively, find the 31^stterm.
Solution:
For a given A.P.,
Number of terms, n = 60
First term, a = 7
Last term, l = 125
⇒ t₆₀ = 125
⇒ a + 59d = 125
⇒ 7 + 59d = 125
⇒ 59d = 118
⇒ d = 2
Hence, t₃₁ = a + 30d = 7 + 30(2) = 7 + 60 = 67

Question 19.
The sum of the 4^th and the 8^th terms of an A.P. is 24 and the sum of the sixth term and the tenth term is 34. Find the first three terms of the A.P.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
t₄ + t₈ = 24 (given)
⇒ (a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12 ….(i)
And,
t₆ + t₁₀ = 34 (given)
⇒ (a + 5d) + (a + 9d) = 34
⇒ 2a + 14d = 34
⇒ a + 7d = 17 ….(ii)
Subtracting (i) from (ii), we get
2d = 5

Question 20.
If the third term of an A.P. is 5 and the seventh terms is 9, find the 17^th term.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Now, t₃ = 5 (given)
⇒ a + 2d = 5 ….(i)
And,
t₇ = 9 (given)
⇒ a + 6d = 9 ….(ii)
Subtracting (i) from (ii), we get
4d = 4
⇒ d = 1
⇒ a + 2(1) = 5
⇒ a = 3
Hence, 17^th term = t₁₇ = a + 16d = 3 + 16(1) = 19

Arithmetic Progression Exercise 10B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In an A.P., ten times of its tenth term is equal to thirty times of its 30th term. Find its 40th term.
Solution:
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Question 2.
How many two-digit numbers are divisible by 3?
Solution:
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Question 3.
Which term of A.P. 5, 15, 25 ………… will be 130 more than its 31st term?
Solution:
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Question 4.
Find the value of p, if x, 2x + p and 3x + 6 are in A.P
Solution:
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Question 5.
If the 3rd and the 9th terms of an arithmetic progression are 4 and -8 respectively, Which term of it is zero?
Solution:
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Question 6.
How many three-digit numbers are divisible by 87?
Solution:
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Question 7.
For what value of n, the nth term of A.P 63, 65, 67, …….. and nth term of A.P. 3, 10, 17,…….. are equal to each other?
Solution:
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Question 8.
Determine the A.P. Whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
Solution:
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Question 9.
If numbers n – 2, 4n – 1 and 5n + 2 are in A.P. find the value of n and its next two terms.
Solution:
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Question 10.

Solution:
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Question 11.
If a, b and c are in A.P show that:
(i) 4a, 4b and 4c are in A.P
(ii) a + 4, b + 4 and c + 4 are in A.P.
Solution:
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Question 12.
An A.P consists of 57 terms of which 7th term is 13 and the last term is 108. Find the 45th term of this A.P.
Solution:
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Question 13.
4th term of an A.P is equal to 3 times its first term and 7th term exceeds twice the 3rd time by I. Find the first term and the common difference.
Solution:
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Question 14.
The sum of the 2nd term and the 7th term of an A.P is 30. If its 15th term is 1 less than twice of its 8th term, find the A.P
Solution:
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Question 15.
In an A.P, if mth term is n and nth term is m, show that its rth term is (m + n – r)
Solution:
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Question 16.
Which term of the A.P 3, 10, 17, ………. Will be 84 more than its 13th term?
Solution:
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Arithmetic Progression Exercise 10C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the sum of the first 22 terms of the A.P.: 8, 3, -2, ………..
Solution:
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Question 2.
How many terms of the A.P. :
24, 21, 18, ……… must be taken so that their sum is 78?
Solution:
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Question 3.
Find the sum of 28 terms of an A.P. whose nth term is 8n – 5.
Solution:
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Question 4(i).
Find the sum of all odd natural numbers less than 50
Solution:
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Question 4(ii).
Find the sum of first 12 natural numbers each of which is a multiple of 7.
Solution:
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Question 5.
Find the sum of first 51 terms of an A.P. whose 2nd and 3rd terms are 14 and 18 respectively.
Solution:
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Question 6.
The sum of first 7 terms of an A.P is 49 and that of first 17 terms of it is 289. Find the sum of first n terms
Solution:
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Question 7.
The first term of an A.P is 5, the last term is 45 and the sum of its terms is 1000. Find the number of terms and the common difference of the A.P.
Solution:
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Question 8.
Find the sum of all natural numbers between 250 and 1000 which are divisible by 9.
Solution:
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Question 9.
The first and the last terms of an A.P. are 34 and 700 respectively. If the common difference is 18, how many terms are there and what is their sum?
Solution:
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Question 10.
In an A.P, the first term is 25, nth term is -17 and the sum of n terms is 132. Find n and the common difference.
Solution:
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Question 11.
If the 8th term of an A.P is 37 and the 15th term is 15 more than the 12th term, find the A.P. Also, find the sum of first 20 terms of A.P.
Solution:
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Question 12.
Find the sum of all multiples of 7 between 300 and 700.
Solution:
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Question 13.

Solution:
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Question 14.
The fourth term of an A.P. is 11 and the term exceeds twice the fourth term by 5 the A.P and the sum of first 50 terms
Solution:
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Arithmetic Progression Exercise 10D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find three numbers in A.P. whose sum is 24 and whose product is 440.
Solution:
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Question 2.
The sum of three consecutive terms of an A.P. is 21 and the slim of their squares is 165. Find these terms.
Solution:
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Question 3.
The angles of a quadrilateral are in A.P. with common difference 20°. Find its angles.
Solution:
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Question 4.
Divide 96 into four parts which are in A.P. and the ratio between product of their means to product of their extremes is 15 : 7.
Solution:
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Question 5.
Find five numbers in A.P. whose sum is \(12 \frac{1}{2}\) and the ratio of the first to the last terms is 2: 3.
Solution:
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Question 6.
Split 207 into three parts such that these parts are in A.P. and the product of the two smaller parts is 4623.
Solution:
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Question 7.
The sum of three numbers in A.P. is 15 the sum of the squares of the extreme is 58. Find the numbers.
Solution:
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Question 8.
Find four numbers in A.P. whose sum is 20 and the sum of whose squares is 120.
Solution:
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Question 9.
Insert one arithmetic mean between 3 and 13.
Solution:

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Question 10.
The angles of a polygon are in A.P. with common difference 5°. If the smallest angle is 120°, find the number of sides of the polygon.
Solution:
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Question 11.
\(\frac{1}{a}, \frac{1}{b} \text { and } \frac{1}{c}\) are in A.P. Show that : be, ca and ab are also in A.P.
Solution:
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Question 12.
Insert four A.M.s between 14 and -1.
Solution:
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Question 13.
Insert five A.M.s between -12 and 8.
Solution:
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Question 14.
Insert six A.M.s between 15 and -15.
Solution:
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Arithmetic Progression Exercise 10E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Two cars start together in the same direction from the same place. The first car goes at uniform speed of 10 km hr^-1 The second car goes at a speed of 8 km h^-1 in the first hour and thereafter increasing the speed by 0.5 km h^-1 each succeeding hour. After how many hours will the two cars meet?
Solution:
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Question 2.
A sum of ₹ 700 is to be paid to give seven cash prizes to the students of a school for their overall academic performance. If the cost of each prize is ₹ 20 less than its preceding prize; find the value of each of the prizes.
Solution:
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Question 3.
An article can be bought by paying ₹ 28,000 at once or by making 12 monthly instalments. If the first instalment paid is ₹ 3,000 and every other instalment is ₹ 100 less than the previous one, find :
(i) amount of instalment paid in the 9th month
(ii) total amount paid in the instalment scheme.
Solution:
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Question 4.
A manufacturer of TV sets produces 600 units in the third year and 700 units in the 7^th year.
Assuming that the production increases uniformly by a fixed number every year, find :
(i) the production in the first year.
(ii) the production in the 10^th year.
(iii) the total production in 7 years.
Solution:
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Question 5.
Mrs. Gupta repays her total loan of ₹ 1.18,000 by paying instalments every month. If the instalment for the first month is ₹ 1,000 and it increases by ₹ 100 every month, what amount will she pay as the 30^th instalment of loan? What amount of loan she still has to pay after the 30^th instalment?
Solution:
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Question 6.
In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be five times of the class to which the respective section belongs. If there are 1 to 10 classes in the school and each class has three sections, find how many trees were planted by the students?
Solution:
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Arithmetic Progression Exercise 10F – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The 6^th term of an A.P. is 16 and the 14^th term is 32. Determine the 36^th term.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Now, t₆ = 16 (given)
⇒ a + 5d = 16 ….(i)
And,
t₁₄ = 32 (given)
⇒ a + 13d = 32 ….(ii)
Subtracting (i) from (ii), we get
8d = 16
⇒ d = 2
⇒ a + 5(2) = 16
⇒ a = 6
Hence, 36^th term = t₃₆ = a + 35d = 6 + 35(2) = 76

Question 2.
If the third and the 9^th terms of an A.P. term is 12 and the last term is 106. Find the 29^th term of the A.P.
Solution:
For an A.P.,a
t₃ = 4
⇒ a + 2d = 4 … (i)
t₉ = -8
⇒ a + 8d = -8 …. (ii)
Subtracting (i) from (ii), we get
6d = -12
⇒ d = -2
Substituting d = -2 in (i), we get
a = 2(-2) = 4
⇒ a – 4 = 4
⇒ a = 8
⇒ General term = t_n = 8 + (n – 1)(-2)
Let p^th term of this A.P. be 0.
⇒ 8 + (0 – 1) (-2) = 0
⇒ 8 – 2p + 2 = 0
⇒ 10 – 2p = 0
⇒ 2p = 10
⇒ p = 5
Thus, 5^th term of this A.P. is 0.

Question 3.
An A.P. consists of 50 terms of which 3^rd term is 12 and the last term is 106. Find the 29^th term of the A.P.
Solution:
For a given A.P.,
Number of terms, n = 50
3^rd term, t₃ = 12
⇒ a + 2d = 12 ….(i)
Last term, l = 106
⇒ t₅₀ = 106
⇒ a + 49d = 106 ….(ii)
Subtracting (i) from (ii), we get
47d = 94
⇒ d = 2
⇒ a + 2(2) = 12
⇒ a = 8
Hence, t₂₉ = a + 28d = 8 + 28(2) = 8 + 56 = 64

Question 4.
Find the arithmetic mean of :
(i) -5 and 41
(ii) 3x – 2y and 3x + 2y
(iii) (m + n)² and (m – n)²
Solution:
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Question 5.
Find the sum of first 10 terms of the A.P. 4 + 6 + 8 + ………
Solution:
Here,
First term, a = 4
Common difference, d = 6 – 4 = 2
n = 10
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Question 6.
Find the sum of first 20 terms of an A.P. whose first term is 3 and the last term is 60.
Solution:
Here,
First term, a = 3
Last term, l = 57
n = 20
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Question 7.
How many terms of the series 18 + 15 + 12 + ……. when added together will give 45 ?
Solution:
Here, we find that
15 – 18 = 12 – 15 = -3
Thus, the given series is an A.P. with first term 18 and common difference -3.
Let the number of term to be added be ‘n’.
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⇒ 90 = n[36 – 3n + 3]
⇒ 90 = n[39 – 3n]
⇒ 90 = 3n[13 – n]
⇒ 30 = 13n – n²
⇒ n² – 13n + 30 = 0
⇒ n² – 10n – 3n + 30 = 0
⇒ n(n – 10) – 3(n – 10) = 0
⇒ (n – 10)(n – 3) = 0
⇒ n – 10 = 0 or n – 3 = 0
⇒ n = 10 or n = 3
Thus, required number of term to be added is 3 or 10.

Question 8.
The n^th term of a sequence is 8 – 5n. Show that the sequence is an A.P.
Solution:
t_n = 8 – 5n
Replacing n by (n + 1), we get
t_n+1 = 8 – 5(n + 1) = 8 – 5n – 5 = 3 – 5n
Now,
t_n+1 – t_n = (3 – 5n) – (8 – 5n) = -5
Since, (t_n+1 – t₂) is independent of n and is therefore a constant.
Hence, the given sequence is an A.P.

Question 9.
The the general term (n^th term) and 23^rd term of the sequence 3, 1, -1, -3, ……
Solution:
The given sequence is 1, -1, -3, …..
Now,
1 – 3 = -1 – 1 = -3 – (-1) = -2
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = -2.
The general term (n^th term) of an A.P. is given by
t_n = a + (n – 1)d
= 3 + (n – 1)(-2)
= 3 – 2n + 2
= 5 – 2n
Hence, 23^rd term = t₂₃ = 5 – 2(23) = 5 – 46 = -41

Question 10.
Which term of the sequence 3, 8, 13, …….. is 78 ?
Solution:
The given sequence is 3, 8, 13, …..
Now,
8 – 3 = 13 – 8 = 5
Hence, the given sequence is an A.P. with first term a = 3 and common difference d = 5.
Let the n^th term of the given A.P. be 78.
⇒ 78 = 3 + (n – 1)(5)
⇒ 75 = 5n – 5
⇒ 5n = 80
⇒ n = 16
Thus, the 16th term of the given sequence is 78.

Question 11.
Is -150 a term of 11, 8, 5, 2, ……… ?
Solution:
The given sequence is 11, 8, 5, 2, …..
Now,
8 – 11 = 5 – 8 = 2 – 5 = -3
Hence, the given sequence is an A.P. with first term a = 11 and common difference d = -3.
The general term of an A.P. is given by
t_n = a + (n – 1)d
⇒ -150 = 11 + (n – 1)(-5)
⇒ -161 = -5n + 5
⇒ 5n = 166
⇒ n =\(\frac{166}{5}\)
The number of terms cannot be a fraction.
So, clearly, -150 is not a term of the given sequence.

Question 12.
How many two digit numbers are divisible by 3 ?
Solution:
The two-digit numbers divisible by 3 are as follows: 12, 15, 18, 21, …….. 99
Clearly, this forms an A.P. with first term, a = 12
and common difference, d = 3
Last term = n^th term= 99
The general term of an A.P. is given by
t_n = a + (n – 1)d
⇒ 99 – 12 + (n – 1)(3)
⇒ 99 – 12 + 3n-3
⇒ 90 – 3n
⇒ n = 30
Thus, 30 two-digit numbers are divisible by 3.

Question 13.
How many multiples of 4 lie between 10 and 250 ?
Solution:
Numbers between 10 and 250 which are multiple of 4 are as follows: 12, 16, 20, 24,……, 248
Clearly, this forms an A.P. with first term a = 12,
common difference d= 4 and last term l = 248
l – a + (n – 1)d
⇒ 248 – 12 + (n – 1) × 4
⇒ 236 – (n – 1) × 4
⇒ n – 1 = 59
⇒ n = 60
Thus, 60 multiples of 4 lie between 10 and 250.

Question 14.
The sum of the 4^th term and the 8^th term of an A.P. is 24 and the sum of 6^th term and the 10^th term is 44. Find the first three terms of the A.P.
Solution:
Given, t₄ + t₈ = 24
(a + 3d) + (a + 7d) = 24
= 2a+ 10d = 24
> a + 5d = 12 ….(i)
And,
t₆₂ + t₁₀ = 44
= (a + 5d) + (a + 9d) = 44
= 2a+ 14d = 44
= a + 7 = 22 …(ii)
Subtracting (i) from (ii), we get
2d = 10
= d = 5
Substituting value of din (i), we get
a + 5 × 5 = 12
= a + 25 = 12
= a = -13 = 1^st term
a + d = -13 + 5 = -8 = 2^nd term
a + 2d = -13 + 2 × 5 = -13 + 10= -3 = 3^rd term
Hence, the first three terms of an A.P. are – 13,- 8 and -5.

Question 15.
The sum of first 14 terms of an A.P. is 1050 and its 14^th term is 140. Find the 20^th term.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference of the given A.P.
Given,
S₁₄= 1050
\(\frac{14}{2}[2 a+(14-1) d]=1050\)
⇒ 7[2a + 13d] = 1050
⇒ 2a + 13d = 150
⇒ a + 6.5d = 75 ….(i)
And, t₁₄ = 140
⇒ a + 13d = 140 ….(ii)
Subtracting (i) from (ii), we get
6.5d = 65
⇒ d = 10
⇒ a + 13(10) = 140
⇒ a = 10
Thus, 20^th term = t₂₀ = 10 + 19d = 10 + 19(10) = 200

Question 16.
The 25^th term of an A.P. exceeds its 9^th term by 16. Find its common difference.
Solution:
n^th term of an A.P. is given by tn= a + (n – 1) d.
⇒ t₂₅ = a + (25 – 1)d = a + 24d and
t₉ = a + (9 – 1)d = a + 8d
According to the condition in the question, we get
t₂₅ = t₉ + 16
⇒ a + 24d = a + 8d + 16
⇒ 16d = 16
⇒ d = 1

Question 17.
For an A.P., show that:
(m + n)^th term + (m – n)^th term = 2 × m^thterm
Solution:
Let a and d be the first term and common difference respectively.
⇒(m + n)^th term = a + (m + n – 1)d …. (i) and
(m – n)^th term = a + (m – n – 1)d …. (ii)
From (i) + (ii), we get
(m + n)^th term + (m – n)^th term
= a + (m + n – 1)d + a + (m – n – 1)d
= a + md + nd – d + a + md – nd – d
= 2a + 2md – 2d
= 2a + (m – 1)2d
= 2[ a + (m – 1)d]
= 2 × m^th term
Hence proved.

Question 18.
If the n^th term of the A.P. 58, 60, 62,…. is equal to the n^th term of the A.P. -2, 5, 12, …., find the value of n.
Solution:
In the first A.P. 58, 60, 62,….
a = 58 and d = 2
t_n = a + (n – 1)d
⇒ t_n = 58 + (n – 1)2 …. (i)
In the first A.P. -2, 5, 12, ….
a = -2 and d = 7
t_n = a + (n – 1)d
⇒ t_n = -2 + (n – 1)7 …. (ii)
Given that the n^th term of first A.P is equal to the n^th term of the second A.P.
⇒58 + (n – 1)2 = -2 + (n – 1)7 … from (i) and (ii)
⇒58 + 2n – 2 = -2 + 7n – 7
⇒ 65 = 5n
⇒ n = 15

Question 19.
Which term of the A.P. 105, 101, 97 … is the first negative term?
Solution:
Here a = 105 and d = 101 – 105 = -4
Let an be the first negative term.
⇒ a_2n < 0
⇒ a + (n – 1)d < 0
⇒ 105 + (n – 1)(-4)

Question 20.
How many three digit numbers are divisible by 7?
Solution:
The first three digit number which is divisible by 7 is 105 and the last digit which is divisible by 7 is 994.
This is an A.P. in which a = 105, d = 7 and t_n = 994.
We know that n^th term of A.P is given by
t_n = a + (n – 1)d.
⇒ 994 = 105 + (n – 1)7
⇒ 889 = 7n – 7
⇒ 896 = 7n
⇒ n = 128
∴ There are 128 three digit numbers which are divisible by 7.

Question 21.
Divide 216 into three parts which are in A.P. and the product of the two smaller parts is 5040.
Solution:
Let the three parts of 216 in A.P be (a – d), a, (a + d).
⇒a – d + a + a + d = 216
⇒ 3a = 216
⇒ a = 72
Given that the product of the two smaller parts is 5040.
⇒ a(a – d ) = 5040
⇒ 72(72 – d) = 5040
⇒ 72 – d = 70
⇒ d = 2
∴ a – d = 72 – 2 = 70, a = 72 and a + d = 72 + 2 = 74
Therefore the three parts of 216 are 70, 72 and 74.

Question 22.
Can 2n² – 7 be the n^th term of an A.P? Explain.
Solution:
We have 2n² – 7,
Substitute n = 1, 2, 3, … , we get
2(1)² – 7, 2(2)² – 7, 2(3)² – 7, 2(4)² – 7, ….
-5, 1, 11, ….
Difference between the first and second term = 1 – (-5) = 6
And Difference between the second and third term = 11 – 1 = 10
Here, the common difference is not same.
Therefore the nth term of an A.P can’t be 2n² – 7.

Question 23.
Find the sum of the A.P., 14, 21, 28, …, 168.
Solution:
Here a = 14 , d = 7 and t_n = 168
t_n = a + (n – 1)d
⇒ 168 = 14 + (n – 1)7
⇒ 154 = 7n – 7
⇒ 154 = 7n – 7
⇒ 161 = 7n
⇒ n = 23
We know that,
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 81
Therefore the sum of the A.P., 14, 21, 28, …, 168 is 2093.

Question 24.
The first term of an A.P. is 20 and the sum of its first seven terms is 2100; find the 31^st term of this A.P.
Solution:
Here a = 20 and S₇ = 2100
We know that,
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 82
To find: t₃₁ =?
t_n = a + (n – 1)d

Therefore the 31st term of the given A.P. is 2820.

Question 25.
Find the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58.
Solution:
First we will reverse the given A.P. as we have to find the sum of last 8 terms of the A.P.
58, …., -8, -10, -12.
Here a = 58 , d = -2
Selina Concise Mathematics Class 10 ICSE Solutions Arithmetic Progression image - 84
Therefore the sum of last 8 terms of the A.P. -12, -10, -8, ……, 58 is 408.

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Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords

November 10, 2023May 26, 2022 by Veerendra

Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 18 Tangents and Intersecting Chords

Tangents and Intersecting Chords Exercise 18A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
The radius of a circle is 8 cm. Calculate the length of a tangent drawn to this circle from a point at a distance of 10 cm from its centre?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 1

Question 2.

In the given figure, O is the centre of the circle and AB is a tangent to the circle at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 2
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 3

Question 3.
Two circles touch each other externally at point P. Q is a point on the common tangent through P. Prove that the tangents QA and QB are equal.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 4
Solution:

Question 4.
Two circles touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 6

Question 5.
Two circles of radii 5 cm and 3 cm are concentric. Calculate the length of a chord of the outer circle which touches the inner.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 7

Question 6.
Three circles touch each other externally. A triangle is formed when the centers of these circles are joined together. Find the radii of the circles, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 8

Question 7.
If the sides of a quadrilateral ABCD touch a circle, prove that AB + CD = BC + AD.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 9
Solution:

Question 8.
If the sides of a parallelogram touch a circle, prove that the parallelogram is a rhombus.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 11
Solution:

From A, AP and AS are tangents to the circle.
Therefore, AP = AS…….(i)
Similarly, we can prove that:
BP = BQ ………(ii)
CR = CQ ………(iii)
DR = DS ………(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
But AB = CD and BC = AD…….(v) Opposite sides of a ||gm
Therefore, AB + AB = BC + BC
2AB = 2 BC
AB = BC ……..(vi)
From (v) and (vi)
AB = BC = CD = DA
Hence, ABCD is a rhombus.

Question 9.
From the given figure prove that:
AP + BQ + CR = BP + CQ + AR.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 13
Also, show that AP + BQ + CR = \(\frac{1}{2}\) × perimeter of triangle ABC.
Solution:

Question 10.
In the figure, if AB = AC then prove that BQ = CQ.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 15
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 16

Question 11.
Radii of two circles are 6.3 cm and 3.6 cm. State the distance between their centers if –
i) they touch each other externally.
ii) they touch each other internally.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 17

Question 12.
From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
i) ∠AOP = ∠BOP
ii) OP is the perpendicular bisector of chord AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 18

Question 13.
In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 19
i) tangent at point P bisects AB.
ii) Angle APB = 90°
Solution:

Question 14.
Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. Prove that:
∠PAQ = 2∠OPQ
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 21

Question 15.
Two parallel tangents of a circle meet a third tangent at point P and Q. Prove that PQ subtends a right angle at the centre.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 22

Question 16.
ABC is a right angled triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 24
Calculate the value of x, the radius of the inscribed circle.
Solution:

Question 17.
In a triangle ABC, the incircle (centre O) touches BC, CA and AB at points P, Q and R respectively. Calculate:
i) ∠QOR
ii) ∠QPR
given that ∠A = 60°
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 26

Question 18.
In the following figure, PQ and PR are tangents to the circle, with centre O. If , calculate:
i) ∠QOR
ii) ∠OQR
iii) ∠QSR
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 27
Solution:

Question 19.
In the given figure, AB is a diameter of the circle, with centre O, and AT is a tangent. Calculate the numerical value of x.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 29
Solution:

Question 20.
In quadrilateral ABCD, angle D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm. Find the radius of the circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 31

Question 21.
In the given figure, PT touches the circle with centre O at point R. Diameter SQ is produced to meet the tangent TR at P.
Given and ∠SPR = x° and ∠QRP = y°
Prove that -;
i) ∠ORS = y°
ii) write an expression connecting x and y
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 32
Solution:

Question 22.
PT is a tangent to the circle at T. If ; calculate:
i) ∠CBT
ii) ∠BAT
iii) ∠APT
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 34
Solution:

Question 23.
In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 36
Solution:

Question 24.
In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 38
Solution:
∠CAB = ∠BAQ = 30°……(AB is angle bisector of ∠CAQ)
∠CAQ = 2∠BAQ = 60°……(AB is angle bisector of ∠CAQ)
∠CAQ + ∠PAC = 180°……(angles in linear pair)
∴∠PAC = 120°
∠PAC = 2∠CAD……(AD is angle bisector of ∠PAC)
∠CAD = 60°

Now,
∠CAD + ∠CAB = 60 + 30 = 90°
∠DAB = 90°
Thus, BD subtends 90° on the circle
So, BD is the diameter of circle

Tangents and Intersecting Chords Exercise 18B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
i) In the given figure, 3 x CP = PD = 9 cm and AP = 4.5 cm. Find BP.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 39
ii) In the given figure, 5 x PA = 3 x AB = 30 cm and PC = 4cm. Find CD.

iii) In the given figure, tangent PT = 12.5 cm and PA = 10 cm; find AB.

Solution:

Question 2.
In the given figure, diameter AB and chord CD of a circle meet at P. PT is a tangent to the circle at T. CD = 7.8 cm, PD = 5 cm, PB = 4 cm. Find
(i) AB.
(ii) the length of tangent PT.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 43
Solution:

Question 3.
In the following figure, PQ is the tangent to the circle at A, DB is a diameter and O is the centre of the circle. If ; ∠ADB = 30° and ∠CBD = 60° calculate:
i) ∠QAD
ii) ∠PAD
iii) ∠CDB
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 45
Solution:

Question 4.
If PQ is a tangent to the circle at R; calculate:
i) ∠PRS
ii) ∠ROT
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 47
Given: O is the centre of the circle and ∠TRQ = 30°
Solution:

Question 5.
AB is diameter and AC is a chord of a circle with centre O such that angle BAC=30º. The tangent to the circle at C intersects AB produced in D. Show that BC = BD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 49

Question 6.
Tangent at P to the circumcircle of triangle PQR is drawn. If this tangent is parallel to side QR, show that triangle PQR is isosceles.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 50

Question 7.
Two circles with centers O and O’ are drawn to intersect each other at points A and B.
Centre O of one circle lies on the circumference of the other circle and CD is drawn tangent to the circle with centre O’ at A. Prove that OA bisects angle BAC.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 51
Solution:

Question 8.
Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that: ∠CPA = ∠DPB
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 54
Solution:

Question 9.
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 56

Question 10.
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If angle BCG = 108° and O is the centre of the circle, find:
i) angle BCT
ii) angle DOC
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 57
Solution:

Question 11.
Two circles intersect each other at point A and B. A straight line PAQ cuts the circle at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T are concyclic.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 59

Question 12.
In the figure, PA is a tangent to the circle. PBC is a secant and AD bisects angle BAC.
Show that the triangle PAD is an isosceles triangle. Also show that:
∠CAD = \(\frac{1}{2}\)(∠PBA – ∠PAB)
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 60
Solution:

Question 13.
Two circles intersect each other at point A and B. Their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 62
Solution:

Question 14.
In the figure, chords AE and BC intersect each other at point D.
i) if , ∠CDE = 90° AB = 5 cm, BD = 4 cm and CD = 9 cm; find DE
ii) If AD = BD, Show that AE = BC.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 64
Solution:

Question 15.
Circles with centers P and Q intersect at points A and B as shown in the figure. CBD is a line segment and EBM is tangent to the circle, with centre Q, at point B. If the circles are congruent; show that CE = BD.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 66
Solution:

Question 16.
In the adjoining figure, O is the centre of the circle and AB is a tangent to it at point B. Find ∠BDC = 65. Find ∠BAO
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 68
Solution:

Tangents and Intersecting Chords Exercise 18C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Prove that of any two chord of a circle, the greater chord is nearer to the centre.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 70

Question 2.
OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
i) If the radius of the circle is 10 cm, find the area of the rhombus.
ii) If the area of the rhombus is \(32 \sqrt{3}\) cm², find the radius of the circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 70

Question 3.
Two circles with centers A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 73

Question 4.
Two chords AB and AC of a circle are equal. Prove that the centre of the circle, lies on the bisector of the angle BAC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 74

Question 5.
The diameter and a chord of circle have a common end-point. If the length of the diameter is 20 cm and the length of the chord is 12 cm, how far is the chord from the centre of the circle?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 75

Question 6.
ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 76

Question 7.
In the given figure, C and D are points on the semicircle described on AB as diameter.
Given angle BAD = 70° and angle DBC = 30°, calculate angle BDC
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 77
Solution:

Question 8.
In cyclic quadrilateral ABCD, ∠A = 3 ∠C and ∠D = 5∠B. Find the measure of each angle of the quadrilateral.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 79

ABCD is a cyclic quadrilateral.
∴ ∠A + ∠C = 180°
⇒ 3∠C + ∠C = 180°
⇒ 4∠C = 180°
⇒ ∠C = 45°

∵ ∠A = 3∠C
⇒ ∠A = 3 × 45°
⇒ ∠A = 135°
Similarly,

∴ ∠B+ ∠D = 180°
⇒∠B + 5∠B = 180°
⇒ 6∠B = 180°
⇒ ∠B = 30°

∵∠D = 5∠B
⇒ ∠D = 5 × 30° >
⇒ ∠D = 150°
Hence, ∠A = 1350, ∠B = 30°, ∠C = 450, ∠D = 150°

Question 9.
Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 80

Question 10.
Bisectors of vertex A, B and C of a triangle ABC intersect its circumcircle at points D, E and F respectively. Prove that angle EDF = \(90^{\circ}-\frac{1}{2} \angle A\)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 81

Question 11.
In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20°, find angle AOD.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 83
Solution:

Question 12.
Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 85

Question 13.
P is the midpoint of an arc APB of a circle. Prove that the tangent drawn at P will be parallel to the chord AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 86

Question 14.
In the given figure, MN is the common chord of two intersecting circles and AB is their common tangent.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 87
Prove that the line NM produced bisects AB at P.
Solution:

Question 15.
In the given figure, ABCD is a cyclic quadrilateral, PQ is tangent to the circle at point C and BD is its diameter. If ∠DCQ = 40° and ∠ABD = 60°, find:
i) ∠DBC
ii) ∠ BCP
iii) ∠ ADB
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 89
Solution:

Question 16.
The given figure shows a circle with centre O and BCD is a tangent to it at C. Show that: ∠ACD + ∠BAC = 90°
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 91
Solution:

Question 17.
ABC is a right triangle with angle B = 90º. A circle with BC as diameter meets by hypotenuse AC at point D.
Prove that –
i) AC × AD = AB²ii) BD²= AD × DC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 93

Question 18.
In the given figure AC = AE.
Show that:
i) CP = EP
ii) BP = DP
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 94
Solution:

Question 19.
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC=120°
Calculate:
i) ∠BEC
ii) ∠ BED
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 96

Question 20.
In the given figure, O is the centre of the circle. Tangents at A and B meet at C. If angle ACO = 30°, find:
(i) angle BCO
(ii) angle AOB
(iii) angle APB
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 97
Solution:

Question 21.
ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6cm (not drawn to scale). Three circles are drawn touching each other with the vertices as their centres. Find the radii of the three circles.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 99
Solution:

Question 22.
In a square ABCD, its diagonal AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and bisector of angle ABD meets AC at N and AM at L. Show that –
i) ∠ONL + ∠OML = 180°
ii) ∠BAM = ∠BMA
iii) ALOB is a cyclic quadrilateral.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 101

Question 23.
The given figure shows a semicircle with centre O and diameter PQ. If PA = AB and ∠BOQ = 140°; find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 102
Solution:

Question 24.
The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°.
Calculate –
i) angle QTR
ii) angle QRP
iii) angle QRS
iv) angle STR
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 105
Solution:

Question 25.
In the given figure, PAT is tangent to the circle with centre O, at point A on its circumference and is parallel to chord BC. If CDQ is a line segment, show that:
i) ∠BAP = ∠ADQ
ii) ∠AOB = 2∠ADQ
(iii) ∠ADQ = ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 108
Solution:

Question 26.
AB is a line segment and M is its midpoint. Three semicircles are drawn with AM, MB and AB as diameters on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semicircles. Show that: AB = 6 x r
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 110

Question 27.
TA and TB are tangents to a circle with centre O from an external point T. OT intersects the circle at point P. Prove that AP bisects the angle TAB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 112

Question 28.
Two circles intersect in points P and Q. A secant passing through P intersects the circle in A and B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 113

Question 29.
Prove that any four vertices of a regular pentagon are concyclic (lie on the same circle)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 114

Question 30.
Chords AB and CD of a circle when extended meet at point X. Given AB = 4 cm, BX = 6 cm and XD = 5 cm. Calculate the length of CD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 115

Question 31.
In the given figure, find TP if AT = 16 cm and AB = 12 cm.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 116
Solution:

Question 32.
In the following figure, A circle is inscribed in the quadrilateral ABCD.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 118
If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.
Solution:

Question 33.
In the figure, XY is the diameter of the circle, PQ is the tangent to the circle at Y. Given that ∠AXB = 50° and ∠ABX = 70°. Calculate ∠BAY and ∠APY.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 120
Solution:

Question 34.
In the given figure, QAP is the tangent at point A and PBD is a straight line. If ∠ACB = 36° and ∠APB = 42°; find:
i) ∠BAP
ii) ∠ABD
iii) ∠QAD
iv) ∠BCD
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 122
Solution:

Question 35.
In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 124
If
∠CAB = 34°, find
i) ∠CBA
ii) ∠CQB
Solution:

Question 36.
In the given figure, O is the centre of the circle. The tangets at B and D intersect each other at point P.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 126
If AB is parallel to CD and ∠ABC = 55°, find:
i) ∠BOD
ii) ∠BPD
Solution:

Question 37.
In the figure given below PQ =QR, ∠RQP = 68°, PC and CQ are tangents to the circle with centre O. Calculate the values of:
i) ∠QOP
ii) ∠QCP
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 128
Solution:

Question 38.
In two concentric circles prove that all chords of the outer circle, which touch the inner circle, are of equal length.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 130

Question 39.
In the figure, given below, AC is a transverse common tangent to two circles with centers P and Q and of radii 6 cm and 3 cm respectively.
Given that AB = 8 cm, calculate PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 131
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 132

Question 40.
In the figure given below, O is the centre of the circum circle of triangle XYZ. Tangents at X and Y intersect at point T. Given ∠XTY = 80° and ∠XOZ = 140°, calculate the value of ∠ZXY.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 133
Solution:

Question 41.

In the given figure, AE and BC intersect each other at point D. If ∠CDE=90°, AB = 5 cm, BD = 4 cm and CD = 9 cm, find AE.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 135
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 136

Question 42.
In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 137
Solution:

Question 43.
In the figure given below, O is the centre of the circle and SP is a tangent. If ∠SRT = 65°, find the values of x, y and z.
Selina Concise Mathematics Class 10 ICSE Solutions Tangents and Intersecting Chords - 139
Solution:

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Selina Concise Mathematics Class 10 ICSE Solutions Circles

November 10, 2023May 26, 2022 by Veerendra

Selina Concise Mathematics Class 10 ICSE Solutions Circles

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 17 Circles

Circles Exercise 17A – Selina Concise Mathematics Class 10 ICSE Solutions

Circles Class 10 Question 1.
In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 1
Solution:

Question 2.
In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°
(i) Prove that AC is a diameter of the circle.
(ii) Find ∠ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 3
Solution:

Question 3.
Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of:
(i) ∠ OCA,
(ii) ∠OAC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 5
Solution:

Question 4.
In each of the following figures, O is the centre of the circle. Find the values of a, b, and c.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 7
Solution:

Question 5.
In each of the following figures, O is the centre of the circle. Find the value of a, b, c and d.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 9
Solution:

Question 6.
In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O₁ and O₂ are the centres of two circles.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 12
Solution:

Question 7.
In the figure given beow, find :
(i) ∠ BCD,
(ii) ∠ ADC,
(iii) ∠ ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 14
Show steps of your workng.
Solution:

Question 8.
In the given figure, O is centre of the circle. If ∠ AOB = 140° and ∠ OAC = 50°; find :
(i) ∠ ACB,
(ii) ∠ OBC,
(iii) ∠ OAB,
(iv) ∠CBA
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 16
Solution:

Question 9.
Calculate :
(i) ∠ CDB,
(ii) ∠ ABC,
(iii) ∠ ACB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 18
Solution:

Question 10.
In the figure given below, ABCD is a eyclic quadrilateral in which ∠ BAD = 75°; ∠ ABD = 58° and ∠ADC = 77°. Find:
(i) ∠ BDC,
(ii) ∠ BCD,
(iii) ∠ BCA.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 20
Solution:

Question 11.
In the following figure, O is centre of the circle and ∆ ABC is equilateral. Find :
(i) ∠ ADB
(ii) ∠ AEB
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 22
Solution:

Question 12.
Given—∠ CAB = 75° and ∠ CBA = 50°. Find the value of ∠ DAB + ∠ ABD
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 24
Solution:

Question 13.
ABCD is a cyclic quadrilateral in a circle with centre O.
If ∠ ADC = 130°; find ∠ BAC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 26
Solution:

Question 14.
In the figure given below, AOB is a diameter of the circle and ∠ AOC = 110°. Find ∠ BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 28
Solution:

Question 15.
In the following figure, O is centre of the circle,
∠ AOB = 60° and ∠ BDC = 100°.
Find ∠ OBC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 30

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 31

Question 16.
ABCD is a cyclic quadrilateral in which ∠ DAC = 27°; ∠ DBA = 50° and ∠ ADB = 33°.
Calculate :
(i) ∠ DBC,
(ii) ∠ DCB,
(iii) ∠ CAB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 32
Solution:

Question 17.
In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°. Find the number of degrees in:
(i) ∠DCE;
(ii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 34
Solution:

Question 17 (old).
In the figure given below, AB is diameter of the circle whose centre is O. Given that:
∠ ECD = ∠ EDC = 32°.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 36
Show that ∠ COF = ∠ CEF.
Solution:

Question 18.
In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 38
Solution:

Question 19.
In the following figure,
(i) if ∠BAD = 96°, find BCD and
(ii) Prove that AD is parallel to FE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 40
Solution:

Question 20.
Prove that:
(i) the parallelogram, inscribed in a circle, is a rectangle.
(ii) the rhombus, inscribed in a circle, is a square.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 42

Question 21.
In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 44
Solution:

Question 22.
Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 46

Question 23.
The figure given below, shows a circle with centre O. Given: ∠ AOC = a and ∠ ABC = b.
(i) Find the relationship between a and b
(ii) Find the measure of angle OAB, if OABC is a parallelogram.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 47
Solution:

Question 24.
Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the centre O is equal to twice the angle APC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 49

Question 24 (old).
ABCD is a quadrilateral inscribed in a circle having ∠A = 60°; O is the centre of the circle. Show that: ∠OBD + ∠ODB = ∠CBD + ∠CDB
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 50

Question 25.
In the figure given RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°
Calculate:
(i) ∠RNM;
(ii) ∠NRM.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 51
Solution:

Question 26.
In the figure given alongside, AB || CD and O is the centre of the circle. If ∠ ADC = 25°; find the angle AEB. Give reasons in support of your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 53
Solution:

Question 27.
Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at Cand D. Prove that AC is parallel to BD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 55
Solution:

Question 28.
ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 57

Question 29.
AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find:
(i) ∠PRB
(ii) ∠PBR
(iii) ∠BPR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 58
Solution:

Question 30.
In the given figure, SP is the bisector of angle RPT and PQRS is a cyclic quadrilateral. Prove that: SQ = SR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 60
Solution:

Question 31.
In the figure, O is the centre of the circle, ∠AOE = 150°, DAO = 51°. Calculate the sizes of the angles CEB and OCE.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 62
Solution:

Question 32.
In the figure, P and Q are the centres of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 64
Solution:

Question 33.
The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. Given that ∠APB = a°. Calculate, in terms of a°, the value of:
(i) obtuse ∠AOB
(ii) ∠ACB
(iii) ∠ADB.
Give reasons for your answers clearly.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 66
Solution:

Question 34.
In the given figure, O is the centre of the circle and ∠ ABC = 55°. Calculate the values of x and y.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 68
Solution:

Question 35.
In the given figure, A is the centre of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that ∠BCD = 2∠ABE
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 70
Solution:

Question 36.
ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate:
(i) ∠ DAB,
(ii) ∠BDC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 72
Solution:

Question 37.
∠ In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠ EAB = 63°; calculate:
(i) ∠EBA,
(ii) BCD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 74
Solution:

Question 38.
In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°; calculate:
(i) ∠ DAB,
(ii) ∠ DBA,
(iii) ∠ DBC,
(iv) ∠ ADC.
Also, show that the ∆AOD is an equilateral triangle.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 76
Solution:

Question 39.
In the given figure, I is the incentre of the ∆ ABC. Bl when produced meets the circumcirle of ∆ ABC at D. Given ∠BAC = 55° and ∠ ACB = 65°, calculate:
(i) ∠DCA,
(ii) ∠ DAC,
(iii) ∠DCI,
(iv) ∠AIC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 78
Solution:

Question 40.
A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that:
(i) ∠ABC = 2 ∠APQ
(ii) ∠ACB = 2 ∠APR
(iii) ∠QPR = 90° – \(\frac{1}{2}\)BAC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 80
Solution:

Question 40 (old).
The sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E; the sides DA and CB are produced to meet at F. If ∠BEC = 42° and ∠BAD = 98°; calculate:
(i) ∠AFB,
(ii) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 82
Solution:

Question 41.
Calculate the angles x, y and z if: \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 84
Solution:

Question 42.
In the given figure, AB = AC = CD and ∠ADC = 38°. Calculate:
(i) Angle ABC
(ii) Angle BEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 86
Solution:

Question 43.
In the given figure, AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, and r in terms of x.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 88
Solution:

Question 44.
In the given figure, AC is the diameter of circle, centre O. CD and BE are parallel. Angle AOB = 80° and angle ACE = 10°. Calculate:
(i) Angle BEC;
(ii) Angle BCD;
(iii) Angle CED.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 90
Solution:

Question 45.
In the given figure, AE is the diameter of circle. Write down the numerical value of ∠ABC + ∠CDE. Give reasons for your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 92
Solution:

Question 46.
In the given figure, AOC is a diameter and AC is parallel to ED. If ∠CBE = 64°, calculate ∠DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 94
Solution:

Question 47.
Use the given figure to find
(i) ∠BAD
(ii) ∠DQB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 96
Solution:

Question 48.
In the given figure, AOB is a diameter and DC is parallel to AB. If ∠ CAB = x°; find (in terms of x) the values of:
(i) ∠COB
(ii) ∠DOC
(iii) ∠DAC
(iv) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 98
Solution:

Question 49.
In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130°. Find:
(i) ∠DAB
(ii) ∠DBA
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 100
Solution:

Question 50.
In the given figure, PQ is the diameter of the circle whose centre is O. Given ∠ROS = 42°; calculate ∠RTS.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 102
Solution:

Question 51.
In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠PQR = 58°; calculate
(i) ∠RPQ
(ii) ∠STP.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 104
Solution:

Question 52.
AOD = 60°; calculate the numerical values of:
AB is the diameter of the circle with centre O. OD is parallel to BC and ∠AOD = 60°; calculate the numerical values of:
(i) ∠ABD,
(ii) ∠DBC,
(iii) ∠ADC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 106
Solution:

Question 53.
In the given figure, the centre of the small circle lies on the circumference of the bigger circle. If ∠APB = 75° and ∠BCD = 40″; find:
(i) ∠AOB,
(ii) ∠ACB,
(iii) ∠ABD,
(iv) ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 108
Solution:

Question 54.
In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°; find:
(i) ∠BCD,
(ii) ∠ACB.
Hence, show that AC is a diameter.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 110
Solution:

Question 55.
In a cyclic quadrilateral ABCD, ∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5; find each angle of the quadrilateral.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 112

Question 56.
The given figure shows a circle with centre O and ∠ABP = 42°. Calculate the measure of
(i) ∠PQB
(ii) ∠QPB + ∠PBQ
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 195
Solution:

Question 57.
In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other. If ∠ MAD =x and ∠BAC = y.
(i) express ∠AMD in terms of x.
(ii) express ∠ABD in terms of y.
(iii) prove that : x = y
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 197
Solution:

Question 61 (old).
In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that ∠APB = 60°.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 199

Circles Exercise 17B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.
Prove it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 200

Question 2.
In the following figure, AD is the diameter of the circle with centre 0. chords AB, BC and CD are equal. If ∠DEF = 110°, calculate:
(i) ∠ AFE,
(ii) ∠FAB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 201

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 202

Question 3.
If two sides of a cycli-quadrilateral are parallel; prove thet:
(i) its other two side are equal.
(ii) its diagonals are equal.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 203

Question 4.
The given figure show a circle with centre O. also, PQ = QR = RS and ∠PTS = 75°. Calculate:
(i) ∠POS,
(ii) ∠ QOR,
(iii) ∠PQR.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 204
Solution:

Question 5.
In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of:
(i) ∠ AOB,
(ii) ∠ ACB,
(iii) ∠ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 206
Solution:

Question 6.
In a regular pentagon ABCDE, inscribed in a circle; find ratio between angle EDA and angel ADC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 208

Question 7.
In the given figure. AB = BC = CD and ∠ABC = 132°, calculate:
(i) ∠AEB,
(ii) ∠ AED,
(iii) ∠COD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 209
Solution:

Question 8.
In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108°, find:
(i) ∠ CAB,
(ii) ∠ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 211
Solution:

Question 9.
The figure shows a circle with centre O. AB is the side of regular pentagon and AC is the side of regular hexagon. Find the angles of triangle ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 213
Solution:

Question 10.
In the given figire, BD is a side of a regularhexagon, DC is a side of a regular pentagon and AD is adiameter. Calculate:
(i) ∠ ADC
(ii) ∠BAD,
(iii) ∠ABC
(iv) ∠ AEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 215
Solution:

Circles Exercise 17C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
In the given circle with diametre AB, find the value of x.

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 218
Solution:

∠ABD = ∠ACD = 30° (Angle in the same segment)
Now in ∆ ADB,
∠BAD + ∠ADB + ∠DBA = 180° (Angles of a A)
But ∠ADB = 90° (Angle in a semi-circle)
∴ x + 90° + 30° = 180° ⇒ x + 120° = 180°
∴ x= 180° – 120° = 60° Ans.

Question 1.
In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 220
Solution:

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 221

Question 2.
In the given figure, ABC is a triangle in which ∠ BAC = 30° Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose centre is O.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 222
Solution:

Question 3.
Prove that the circle drawn on any one a the equalside of an isoscele triangle as diameter bisects the base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 224

Question 3 (old).
The given figure show two circles with centres A and B; and radii 5 cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 225
Solution:

Question 4.
In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠ CBE = 65°, calculate ∠DEC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 227
Solution:

Question 5.
The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 229

Question 6.
In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(i) ∠BDC
(ii) ∠BEC
(iii) ∠BAC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 230
Solution:

Question 7.
D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Provet that the points B, C, E and D are concyclic.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 232

Question 7 (old).
Chords AB and CD of a circle intersect each other at point P such that AP = CP.
Show that: AB = CD.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 233
Solution:

Question 8.
In the given rigure, ABCD is a cyclic eqadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ ADC = 92°, ∠ FAE = 20°; determine ∠ BCD. Given reason in support of your answer.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 235
Solution:

Question 9.
If I is the incentre of triangle ABC and Al when produced meets the cicrumcircle of triangle ABC in points D. if ∠ BAC = 66° and ∠ = 80o.calculate:
(i) ∠ DBC
(ii) ∠ IBC
(iii) ∠ BIC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 237
Solution:

Question 10.
In the given figure, AB = AD = DC = PB and ∠ DBC = x°. Determine, in terms of x:
(i) ∠ ABD,
(ii) ∠ APB.
Hence or otherwise, prove thet AP is parallel to DB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 240
Solution:

Question 11.
In the given figure; ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠ CQE are supplementary.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 242
Solution:

Question 12.
In the given, AB is the diameter of the circle with centre O.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 244
If ∠ ADC = 32°, find angle BOC.
Solution:

Question 13.
In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ prouduced meet at point A: whereas sides PQ and SR produced meet at point B.
If ∠A: ∠B = 2 : 1;find angles A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 246

Question 17 (old).
If the following figure, AB is the diameter of a circle with centre O and CD is the chord with lengh equal radius OA.
If AC produced and BD produed meet at point p; show that ∠APB = 60°
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 248
Solution:

Question 14.
In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 251
If the bisector of angle A meet BC at point E and the given circle at point F, prove that:
(i) EF = FC
(ii) BF =DF
Solution:

Question 15.
ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point e; whereas sides BC and AD produced meet at point F. I f ∠ DCF : ∠F : ∠E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 253

Question 16.
The following figure shows a cicrcle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, Find the perimetre of the cyclic quadrilateral PORS.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 255
Solution:

Question 17.
In the following figure, AB is the diameter of a circle with centre O. If chord AC = chord AD.prove that:
(i) arc BC = arc DB
(ii) AB is bisector of ∠ CAD.
Further if the lenghof arc AC is twice the lengthof arc BC find :
(a) ∠ BAC
(b) ∠ ABC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 257
Solution:

Question 18.
In cyclic quadrilateral ABCD; AD = BC, ∠ = 30° and ∠ = 70°; find;
(i) ∠ BCD
(ii) ∠BCA
(iii) ∠ABC
(iv) ∠ ADC
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 260

Question 19.
In the given figure, ∠ACE = 43° and ∠ = 62°; find the values of a, b and c.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 262
Solution:

Question 20.
In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠ BAC = 25°
Find
(i) ∠ CAD
(ii) ∠ CBD
(iii) ∠ ADC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 264
Solution:

Question 21.
ABCD is a cyclic quadrilateral of a circle with centre o such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle..if AD and BC produced meet at P, show that APB =60°
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 266

Question 22.
In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 267
Solution:

Question 23.
In the figure, given below, AD = BC, ∠ BAC = 30° and ∠ = 70° find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC
(iv) ∠ADC
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 269
Solution:

Question 24.
In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :
(i) ∠OBD
(ii) ∠AOB
(iii) ∠BED (2016)
Solution:
i. AD is parallel to BC, i.e., OD is parallel to BC and BD is transversal.
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 271

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 272

Selina Concise Mathematics Class 10 ICSE Solutions Circles - 273

Question 25.
In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of
i. ∠BCD
ii. ∠BOD
iii. ∠OBD
Selina Concise Mathematics Class 10 ICSE Solutions Circles - 274
Solution:
∠DAE and ∠DAB are linear pair
So,
∠DAE + ∠DAB = 180°
∴∠DAB = 110°

Also,
∠BCD + ∠DAB = 180°……Opp. Angles of cyclic quadrilateral BADC
∴∠BCD = 70°
∠BCD = \(\frac { 1 }{ 2 }\) ∠BOD…angles subtended by an arc on the centre and on the circle
∴∠BOD = 140°

In ΔBOD,
OB = OD……radii of same circle
So,
∠OBD =∠ODB……isosceles triangle theorem
∠OBD + ∠ODB + ∠BOD = 180°……sum of angles of triangle
2∠OBD = 40°
∠OBD = 20°

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Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions)

November 10, 2023May 26, 2022 by Veerendra

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions)

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 16 Loci (Locus and Its Constructions)

Locus and Its Constructions Exercise 16A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Given— PQ is perpendicular bisector of side AB of the triangle ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 1
Prove— Q is equidistant from A and B.
Solution:

Construction: Join AQ
Proof: In ∆AQP and ∆BQP,
AP = BP (given)
∠QPA = ∠QPB (Each = 90 )
PQ = PQ (Common)
By Side-Angle-Side criterian of congruence, we have
∆AQP ≅ ∆BQP (SAS postulate)
The corresponding parts of the triangle are congruent
∴ AQ = BQ (CPCT)
Hence Q is equidistant from A and B.

Question 2.
Given— CP is bisector of angle C of ∆ ABC.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 3
Prove— P is equidistant from AC and BC.
Solution:

Question 3.
Given— AX bisects angle BAG and PQ is perpendicular bisector of AC which meets AX at point Y.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 5
Prove—
(i) X is equidistant from AB and AC.
(ii) Y is equidistant from A and C.
Solution:

Question 4.
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
Solution:
Given: In triangle ABC, AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 7
Steps of Construction:
i) Draw a line segment BC = 6.3 cm
ii) With centre B and radius 4.2 cm, draw an arc.
iii) With centre C and radius 5 cm, draw another arc which intersects the first arc at A.
iv) Join AB and AC.
∆ABC is the required triangle.
v) Again with centre B and C and radius greater than \(\frac{1}{2} \mathrm{BC}\) draw arcs which intersects each other at L and M.
vi) Join LM intersecting AC at D and BC at E.
vii) Join DB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) 12
Hence, D is equidistant from B and C.

Question 5.
In each of the given figures: PA = PH and QA = QB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) 4
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of points equidistant from two given fixed points.
Solution:
Construction: Join PQ which meets AB in D.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 9
Proof: P is equidistant from A and B.
∴ P lies on the perpendicular bisector of AB.
Similarly, Q is equidistant from A and B.
∴ Q lies on perpendicular bisector of AB.
∴ P and Q both lie on the perpendicular bisector of AB.
∴ PQ is perpendicular bisector of AB.
Hence, locus of the points which are equidistant from two fixed points, is a perpendicular bisector of the line joining the fixed points.

Question 6.
Construct a right angled triangle PQR, in which ∠ Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 10
Steps of Construction:
i) Draw a line segment QR = 4.5 cm
ii) At Q, draw a ray QX making an angle of 90°
iii) With centre R and radius 8 cm, draw an arc which intersects QX at P.
iv) Join RP.
∆PQR is the required triangle.
v) Draw the bisector of ∠PQR which meets PR in T.
vi) From T, draw perpendicular PL and PM respectively on PQ and QR.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 11
Hence, T is equidistant from PQ and QR.

Question 7.
Construct a triangle ABC in which angle ABC = 75°. AB = 5 cm and BC = 6.4 cm. Draw perpendicular bisector of side BC and also the bisector of angle ACB. If these bisectors intersect each other at point P ; prove that P is equidistant from B and C ; and also from AC and BC.
Hence P is equidistant from AC and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 12

Question 8.
In parallelogram ABCD, side AB is greater than side BC and P is a point in AC such that PB bisects angle B. Prove that P is equidistant from AB and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 14

Question 9.
In triangle LMN, bisectors of interior angles at L and N intersect each other at point A.
Prove that –
(i) point A is equidistant from all the three sides of the triangle.
(ii) AM bisects angle LMN.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 15
Construction: Join AM
Proof:
∵ A lies on bisector of ∠N
∴A is equidistant from MN and LN.
Again, A lies on bisector of ∠L
∴ A is equidistant from LN and LM.
Hence, A is equidistant from all sides of the triangle LMN.
∴ A lies on the bisector of ∠M

Question 10.
Use ruler and compasses only for this question:
(i) construct ∆ABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
(ii) Construct the locus of points inside the triangle which are equidistant from BA and BC.
(iii) Construct the locus of points inside the triangle which are equidistant from B and C.
(iv) Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
Solution:
Steps of construction:
(i) Draw line BC = 6 cm and an angle CBX = 60o. Cut off AB = 3.5. Join AC, triangle ABC is the required triangle.
(ii) Draw perpendicular bisector of BC and bisector of angle B
(iii) Bisector of angle B meets bisector of BC at P.
⇒ BP is the required length, where, PB = 3.5 cm
(iv) P is the point which is equidistant from BA and BC, also equidistant from B and C.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 16
PB=3.6 cm

Question 11.
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point E Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 17
Prove that:
(i) F is equidistant from A and B.
(ii) F is equidistant from AB and AC.
Solution:

Question 12.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 21
Since P lies on the bisector of angle B,
therefore, P is equidistant from AB and BC …. (1)
Similarly, P lies on the bisector of angle C,
therefore, P is equidistant from BC and CD …. (2)
From (1) and (2),
Hence, P is equidistant from AB and CD.

Question 13.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 22
Steps of construction:
(i) Draw a line segment AB of 6 cm.
(ii) Draw perpendicular bisector LM of AB. LM is the required locus.
(iii) Take any point on LM say P.
(iv) Join PA and PB.
Since, P lies on the right bisector of line AB.
Therefore, P is equidistant from A and B.
i.e. PA = PB
Hence, Perpendicular bisector of AB is the locus of all points which are equidistant from A and B.

Question 14.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 23
Steps of Construction:
i) Draw a ray BC.
ii) Construct a ray RA making an angle of 75° with BC. Therefore, ABC= 75°.
iii) Draw the angle bisector BP of ∠ABC.
BP is the required locus.
iv) Take any point D on BP.
v) From D, draw DE ⊥ AB and DF ⊥ BC
Since D lies on the angle bisector BP of ∠ABC
D is equidistant from AB and BC.
Hence, DE = DF
Similarly, any point on BP is equidistant from AB and BC.
Therefore, BP is the locus of all points which are equidistant from AB and BC.

Question 15.
Draw an angle ABC = 60°, having AB = 4.6 cm and BC = 5 cm. Find a point P equidistant from AB and BC ; and also equidistant from A and B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 24
Steps of Construction:
i) Draw a line segment BC = 5 cm
ii) At B, draw a ray BX making an angle of 60° and cut off BA = 4.6 cm.
iii) Draw the angle bisector of ∠ABC.
iv) Draw the perpendicular bisector of AB which intersects the angle bisector at P.
P is the required point which is equidistant from AB and BC, as well as from A and B.

Question 16.
In the figure given below, find a point P on CD equidistant from points A and B.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 25
Solution:

Steps of Construction:
i) AB and CD are the two lines given.
ii) Draw a perpendicular bisector of line AB which intersects CD in P.
P is the required point which is equidistant from A and B.
Since P lies on perpendicular bisector of AB; PA = PB.

Question 17.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 27
Solution:

Steps of Construction:
i) In the given triangle, draw the angle bisector of ∠BAC.
ii) Draw the perpendicular bisector of BC which intersects the angle bisector at P.
P is the required point which is equidistant from AB and AC as well as from B and C.
Since P lies on angle bisector of ∠BAC,
It is equidistant from AB and AC.
Again, P lies on perpendicular bisector of BC,
Therefore, it is equidistant from B and C.

Question 18.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ ABC = 60°. Locate by construction the point P such that :
(i) P is equidistant from B and C.
(ii) P is equidistant from AB and BC.
(iii) Measure and record the length of PB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 29
Solution:
Steps of Construction:
1) Draw a line segment AB = 7 cm.
2) Draw angle ∠ABC = 60° with the help of compass.
3) Cut off BC = 8 cm.
4) Join A and C.
5) The triangle ABC so formed is the required triangle.
i) Draw the perpendicular bisector of BC. The point situated on this line will be equidistant from B and C.
ii) Draw the angle bisector of ∠ABC. Any point situated on this angular bisector is equidistant from lines AB and BC.
The point which fulfills the condition required in (i) and (ii) is the intersection point of bisector of line BC and angular bisector of ∠ABC.
P is the required point which is equidistant from AB and AC as well as from B and C.
On measuring the length of line segment PB, it is equal to 4.5 cm.

Question 19.
On a graph paper, draw lines x = 3 and y = -5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 30
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 3 which is parallel to y-axis
And draw another line m, y = -5, which is parallel to x-axis
These two lines intersect each other at P.
Now draw the angle bisector p of angle P.
Since p is the angle bisector of P, any point on P is equidistant from l and m.
Therefore, this line p is equidistant from l and m.

Question 20.
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distance from the given line is always equal to 3 units.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 31
On the graph, draw axis XOX’ and YOY’
Draw a line l, x = 6 which is parallel to y-axis
Take points P and Q which are at a distance of 3 units from the line l.
Draw lines m and n from P and Q parallel to l
With locus = 3, two lines can be drawn x = 3 and x = 9.

Locus and Its Constructions Exercise 16B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Describe the locus of a point at a distance of 3 cm from a fixed point.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 32
The locus of a point which is 3 cm away from a fixed point is circumference of a circle whose radius is 3 cm and the fixed point is the centre of the circle.

Question 2.
Describe the locus of a point at a distance of 2 cm from a fixed line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 33
The locus of a point at a distance of 2 cm from a fixed line AB is a pair of straight lines l and m which are parallel to the given line at a distance of 2 cm.

Question 3.
Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 34
The locus of the centre of a wheel, which is going straight along a level road will be a straight line parallel to the road at a distance equal to the radius of the wheel.

Question 4.
Describe the locus of the moving end of the minute hand of a clock.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 35
The locus of the moving end of the minute hand of the clock will be a circle where radius will be the length of the minute hand.

Question 5.
Describe the locus of a stone dropped from the top of a tower.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 36
The locus of a stone which is dropped from the top of a tower will be a vertical line through the point from which the stone is dropped.

Question 6.
Describe the locus of a runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) 31
The locus of the runner, running around a circular track and always keeping a distance of 1.5 m from the inner edge will be the circumference of a circle whose radius is equal to the radius of the inner circular track plus 1.5 m.

Question 7.
Describe the locus of the door handle as the door opens.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 38
The locus of the door handle will be the circumference of a circle with centre at the axis of rotation of the door and radius equal to the distance between the door handle and the axis of rotation of the door.

Question 8.
Describe the locus of a point inside a circle and equidistant from two fixed points on the circumference of the circle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 39
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.

Question 9.
Describe the locus of the centers of all circles passing through two fixed points.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 40
The locus of the centre of all the circles which pass through two fixed points will be the perpendicular bisector of the line segment joining the two given fixed points.

Question 10.
Describe the locus of vertices of all isosceles triangles having a common base.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 41
The locus of vertices of all isosceles triangles having a common base will be the perpendicular bisector of the common base of the triangles.

Question 11.
Describe the locus of a point in space which is always at a distance of 4 cm from a fixed point.
Solution:
The locus of a point in space is the surface of the sphere whose centre is the fixed point and radius equal to 4 cm.

Question 12.
Describe the locus of a point P so that:
AB² = AP² + BP², where A and B are two fixed points.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 42
The locus of the point P is the circumference of a circle with AB as diameter and satisfies the condition AB² = AP² + BP².

Question 13.
Describe the locus of a point in rhombus ABCD, so that it is equidistant from
i) AB and BC
ii) B and D.
Solution:
i)
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 43
The locus of the point in a rhombus ABCD which is equidistant from AB and BC will be the diagonal BD.
ii)

The locus of the point in a rhombus ABCD which is equidistant from B and D will be the diagonal AC.

Question 14.
The speed of sound is 332 meters per second. A gun is fired. Describe the locus of all the people on the Earth’s surface, who hear the sound exactly one second later.
Solution:
The locus of all the people on Earth’s surface is the circumference of a circle whose radius is 332 m and centre is the point where the gun is fired.

Question 15.
Describe:
i) The locus of points at distances less than 3 cm from a given point.
ii) The locus of points at distances greater than 4 cm from a given point.
iii) The locus of points at distances less than or equal to 2.5 cm from a given point.
iv) The locus of points at distances greater than or equal to 35 mm from a given point.
v)The locus of the centre of a given circle which rolls around the outside of a second circle and is always touching it.
vi) The locus of the centers of all circles that are tangent to both the arms of a given angle.
vii) The locus of the mid-points of all chords parallel to a given chord of a circle.
viii) The locus of points within a circle that are equidistant from the end points of a given chord.
Solution:
i) The locus is the space inside of the circle whose radius is 3 cm and the centre is the fixed point which is given.
ii) The locus is the space outside of the circle whose radius is 4 cm and centre is the fixed point which is given.
iii) The locus is the space inside and circumference of the circle with a radius of 2.5 cm and the centre is the given fixed point.
iv) The locus is the space outside and circumference of the circle with a radius of 35 mm and the centre is the given fixed point.
v) The locus is the circumference of the circle concentric with the second circle whose radius is equal to the sum of the radii of the two given circles.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 45
vi) The locus of the centre of all circles whose tangents are the arms of a given angle is the bisector of that angle.

vii) The locus of the mid-points of the chords which are parallel to a given chords is the diameter perpendicular to the given chords.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 47
viii) The locus of the points within a circle which are equidistant from the end points of a given chord is the diameter which is perpendicular bisector of the given chord.

Question 16.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 49
Draw a line XY parallel to the base BC from the vertex A.
This line is the locus of vertex A of all the triangles which have the base BC and length of altitude equal to AD.

Question 17.
In the given figure, obtain all the points equidistant from lines m and n ; and 2.5 cm from O.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 50
Solution:

Draw an angle bisector PQ and XY of angles formed by the lines m and n. From O, draw arcs with radius 2.5 cm, which intersect the angle bisectors at a, b, c and d respectively.
Hence, a, b, c and d are the required four points.

Question PQ.
By actual drawing obtain the points equidistant from lines m and n and 6 cm from the point P, where P is 2 cm above m, m is parallel to n and m is 6 cm above n.
Solution:
Steps of construction:
i) Draw a linen.
ii) Take a point Lonn and draw a perpendicular to n.
iii) Cut off LM = 6 cm and draw a line q, the perpendicular bisector of LM.
iv) At M, draw a line m making an angle of 90°.
v) Produce LM and mark a point P such that PM = 2 cm.
vi) From P, draw an arc with 6 cm radius which intersects the line q, the perpendicular bisector of LM, at A and B.
A and B are the required points which are equidistant from m and n and are at a distance of 6 cm from P.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 52

Question 18.
A straight line AB is 8 cm long. Draw and describe the locus of a point which is:
(i) always 4 cm from the line AB
(ii) equidistant from A and B.
Mark the two points X and Y, which are 4 cm from AB and equidistant from A and B. Describe the figure AXBY.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 53
(i) Draw a line segment AB = 8 cm.
(ii) Draw two parallel lines l and m to AB at a distance of 4 cm.
(iii) Draw the perpendicular bisector of AB which intersects the parallel lines l and m at X and Y respectively then, X and Y are the required points.
(iv) Join AX, AY, BX and BY.
The figure AXBY is a square as its diagonals are equal and intersect at 90°.

Question 19.
Angle ABC = 60° and BA = BC = 8 cm. The mid-points of BA and BC are M and N respec¬tively. Draw and describe the locus of a point which is :
(i) equidistant from BA and BC.
(ii) 4 cm from M.
(iii) 4 cm from N.
Mark the point P, which is 4 cm from both M and N, and equidistant from BA and BC. Join MP and NP, and describe the figure BMPN.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 54
i) Draw an angle of 60° with AB = BC = 8 cm
ii) Draw the angle bisector BX of ∠ABC
iii) With centre M and N, draw circles of radius equal to 4 cm, which intersects each other at P. P is the required point.
iv) Join MP, NP
BMPN is a rhombus since MP = BM = NB = NP = 4 cm

Question 20.
Draw a triangle ABC in which AB = 6 cm, BC = 4.5 cm and AC = 5 cm. Draw and label:
(i) the locus of the centers of all circles which touch AB and AC.
(ii) the locus of the centers of all circles of radius 2 cm which touch AB.
Hence, construct the circle of radius 2 cm which touches AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 55
Steps of Construction:
i) Draw a line segment BC = 4.5 cm
ii) With B as centre and radius 6 cm and C as centre and radius 5 cm, draw arcs which intersect each other at A.
iii) Join AB and AC.
ABC is the required triangle.
iv) Draw the angle bisector of ∠BAC
v) Draw lines parallel to AB and AC at a distance of 2 cm, which intersect each other and AD at O.
vi) With centre O and radius 2 cm, draw a circle which touches AB and AC.

Question 21.
Construct a triangle ABC, having given AB = 4.8 cm. AC = 4 cm and ∠ A = 75°. Find a point P.
(i) inside the triangle ABC.
(ii) outside the triangle ABC.
equidistant from B and C; and at a distance of 1.2 cm from BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 56
Steps of Construction:
i) Draw a line segment AB = 4.8 cm
ii) At A, draw a ray AX making an angle of 75°
iii) Cut off AC = 4 cm from AX
iv) Join BC.
ABC is the required triangle.
v) Draw two lines l and m parallel to BC at a distance of 1.2 cm
vi) Draw the perpendicular bisector of BC which intersects l and m at P and P’
P and P’ are the required points which are inside and outside the given triangle ABC.

Question PQ.
O is a fixed point. Point P moves along a fixed line AB. Q is a point on OP produced such that OP = PQ. Prove that the locus of point Q is a line parallel to AB.
Solution:
P moves along AB, and Qmoves in such a way that PQ is always equal to OP.
But Pis the mid-point of OQ
Now in ∆OQQ’
P’and P” are the mid-points of OQ’ and OQ”
Therefore, AB||Q’Q”
Therefore, Locus of Q is a line CD which is parallel to AB.
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 57

Question 22.
Draw an angle ABC = 75°. Find a point P such that P is at a distance of 2 cm from AB and 1.5 cm from BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 58
Steps of Construction:
i) Draw a ray BC.
ii) At B, draw a ray BA making an angle of 75° with BC.
iii) Draw a line l parallel to AB at a distance of 2 cm
iv) Draw another line m parallel to BC at a distance of 1.5 cm which intersects line l at P.
P is the required point.

Question 23.
Construct a triangle ABC, with AB = 5.6 cm, AC = BC = 9.2 cm. Find the points equidistant from AB and AC; and also 2 cm from BC. Measure the distance between the two points obtained.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 59
Steps of Construction:
i) Draw a line segment AB = 5.6 cm
ii) From A and B, as centers and radius 9.2 cm, make two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw two lines n and m parallel to BC at a distance of 2 cm
v) Draw the angle bisector of ∠BAC which intersects m and n at P and Q respectively.
P and Q are the required points which are equidistant from AB and AC.
On measuring the distance between P and Q is 4.3 cm.

Question 24.
Construct a triangle ABC, with AB = 6 cm, AC = BC = 9 cm. Find a point 4 cm from A and equidistant from B and C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 60
Steps of Construction:
i) Draw a line segment AB = 6 cm
ii) With A and B as centers and radius 9 cm, draw two arcs which intersect each other at C.
iii) Join AC and BC.
iv) Draw the perpendicular bisector of BC.
v) With A as centre and radius 4 cm, draw an arc which intersects the perpendicular bisector of BC at P.
P is the required point which is equidistant from B and C and at a distance of 4 cm from A.

Question 25.
Ruler and compasses may be used in this question. All construction lines and arcs must be clearly shown and be of sufficient length and clarity to permit assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm and angle ABC = 60°.
(ii) Construct the locus of all points inside triangle ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base and which are equal in area to triangle ABC.
(iv) Mark the point Q, in your construction, which would make A QBC equal in area to A ABC, and isosceles.
(v) Measure and record the length of CQ.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 61
Steps of Construction:
(i) Draw a line segment BC = 6 cm.
(ii) At B, draw a ray BX making an angle 60 degree and cut off BA=9 cm.
(iii) Join AC. ABC is the required triangle.
(iv) Draw perpendicular bisector of BC which intersects BA in M, then any point on LM is equidistant from B and C.
(v) Through A, draw a line m || BC.
(vi) The perpendicular bisector of BC and the parallel line m intersect each other at Q.
(vii) Then triangle QBC is equal in area to triangle ABC. m is the locus of all points through which any triangle with base BC will be equal in area of triangle ABC.
On measuring CQ = 8.4 cm.

Question 26.
State the locus of a point in a rhombus ABCD, which is equidistant
(i) from AB and AD;
(ii) from the vertices A and C.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 62

Question 27.
Use a graph paper for this question. Take 2 cm = 1 unit on both the axes.
(i) Plot the points A(1,1), B(5,3) and C(2,7).
(ii) Construct the locus of points equidistant from A and B.
(iii) Construct the locus of points equidistant from AB and AC.
(iv) Locate the point P such that PA = PB and P is equidistant from AB and AC.
(v) Measure and record the length PA in cm.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 63+
Steps of Construction:
i) Plot the points A(1, 1), B(5, 3) and C(2, 7) on the graph and join AB, BC and CA.
ii) Draw the perpendicular bisector of AB and angle bisector of angle A which intersect each other at P.
P is the required point.
Since P lies on the perpendicular bisector of AB.
Therefore, P is equidistant from A and B.
Again,
Since P lies on the angle bisector of angle A.
Therefore, P is equidistant from AB and AC.
On measuring, the length of PA = 5.2 cm

Question 28.
Construct an isosceles triangle ABC such that AB = 6 cm, BC=AC=4cm. Bisect angle C internally and mark a point P on this bisector such that CP = 5cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 64
Steps of Construction:
i) Draw a line segment AB = 6 cm.
ii) With centers A and B and radius 4 cm, draw two arcs which intersect each other at C.
iii) Join CA and CB.
iv) Draw the angle bisector of angle C and cut off CP = 5 cm.
v) A line m is drawn parallel to AB at a distance of 5 cm.
vi) P as centre and radius 5 cm, draw arcs which intersect the line m at Q and R.
vii) Join PQ, PR and AQ.
Q and R are the required points.

Question PQ.
Use ruler and compasses only for this question. Draw a circle of radius 4 cm and mark two chords AB and AC of the circle of lengths 6 cm and 5 cm respectively.
(i) Construct the locus of points, inside the circle, that are equidistant from A and C. Prove your construction.
(ii) Construct the locus of points, inside the circle, that are equidistant from AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 65

Steps of Construction:
i) Draw a circle with radius = 4 cm.
ii) Take a point A on it.
iii) A as centre and radius 6 cm, draw an arc which intersects the circle at B.
iv) Again A as centre and radius 5 cm, draw an arc which intersects the circle at C.
v) Join AB and AC.
vi) Draw the perpendicular bisector of AC, which intersects AC at Mand meets the circle at E and F.
EF is the locus of points inside the circle which are equidistant from A and C.
vii) Join AE, AF, CE and CF.
Proof:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 66
Similarly, we can prove that CF = AF
Hence EF is the locus of points which are equidistant from A and C.
ii) Draw the bisector of angle A which meets the circle at N.
Therefore. Locus of points inside the circle which are equidistant from AB and AC is the perpendicular bisector of angle A.

Question 29.
Plot the points A(2,9), B(-1,3) and C(6,3) on a graph paper. On the same graph paper, draw the locus of point A so that the area of triangle ABC remains the same as A moves.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 67
Steps of construction:
i) Plot the given points on graph paper.
ii) Join AB, BC and AC.
iii) Draw a line parallel to BC at A and mark it as CD.
CD is the required locus of point A where area of triangle ABC remains same on moving point A.

Question 30.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that:
(a) P is equidistant from A B and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 68
i) Steps of Construction:
1) Draw a line segment BC = 5 cm
2) B as centre and radius 4 cm draw an arc at an angle of 45 degrees from BC.
3) Join PC.
4) B and C as centers, draw two perpendiculars to BC.
5) P as centre and radius PC, cut an arc on the perpendicular on C at D.
6) D as centre, draw a line parallel to BC which intersects the perpendicular on B at A.
ABCD is the required rectangle such that P is equidistant from AB and BC (since BD is angle bisector of angle B) as well as C and D.
ii) On measuring AB = 5.7 cm

Question 31.
Use ruler and compasses only for the following questions. All constructions lines and arcs must be clearly shown.
(i) Construct a ∆ABC in which BC = 6.5 cm, ∠ABC = 60°, AB = 5 cm.
(ii) Construct the locus of points at a distance of 3.5 cm from A.
(iii) Construct the locus of points equidistant from AC and BC.
(iv) Mark 2 points X and Y which are at distance of 3.5 cm from A and also equidistant from AC and BC. Measure XY.
Solution:
i. Steps of construction:

Draw BC = 6.5 cm using a ruler.
With B as center and radius equal to approximately half of BC, draw an arc that cuts the segment BC at Q.
With Q as center and same radius, cut the previous arc at P.
Join BP and extend it.
With B as center and radius 5 cm, draw an arc that cuts the arm PB to obtain point A.
Join AC to obtain ΔABC.

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 69

ii. With A as center and radius 3.5 cm, draw a circle.
The circumference of a circle is the required locus.

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 70
iii. Draw CH, which is bisector of Δ ACB. CH is the required locus.

Selina Concise Mathematics Class 10 ICSE Solutions Loci (Locus and Its Constructions) image - 71
iv. Circle with center A and line CH meet at points X and Y as shown in the figure. xy = 8.2 cm (approximately)

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Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line

November 10, 2023May 26, 2022 by Veerendra

Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line

Selina Publishers Concise Mathematics Class 10 ICSE Solutions Chapter 14 Equation of a Line

Equation of a Line Exercise 14A – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find, which of the following points lie on the line x – 2y + 5 = 0:
(i) (1, 3) (ii) (0, 5)
(iii) (-5, 0) (iv) (5, 5)
(v) (2, -1.5) (vi) (-2, -1.5)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 1

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 2
Solution:

Question 3.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 5

Question 4.
For what value of k will the point (3, -k) lie on the line 9x + 4y = 3?
Solution:
The given equation of the line is 9x + 4y = 3.
Put x = 3 and y = -k, we have:
9(3) + 4(-k) = 3
27 – 4k = 3
4k = 27 – 3 = 24
k = 6

Question 5.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 7

Question 6.
Does the line 3x – 5y = 6 bisect the join of (5, -2) and (-1, 2)?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 8

Question 7.
(i) The line y = 3x – 2 bisects the join of (a, 3) and (2, -5), find the value of a.
(ii) The line x – 6y + 11 = 0 bisects the join of (8, -1) and (0, k). Find the value of k.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 9

Question 8.
(i) The point (-3, 2) lies on the line ax + 3y + 6 = 0, calculate the value of a.
(ii) The line y = mx + 8 contains the point (-4, 4), calculate the value of m.
Solution:
(i) Given, the point (-3, 2) lies on the line ax + 3y + 6 = 0.
Substituting x = -3 and y = 2 in the given equation, we have:
a(-3) + 3(2) + 6 = 0
-3a + 12 = 0
3a = 12
a = 4
(ii) Given, the line y = mx + 8 contains the point (-4, 4).
Substituting x = -4 and y = 4 in the given equation, we have:
4 = -4m + 8
4m = 4
m = 1

Question 9.
The point P divides the join of (2, 1) and (-3, 6) in the ratio 2: 3. Does P lie on the line x – 5y + 15 = 0?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 10

Question 10.
The line segment joining the points (5, -4) and (2, 2) is divided by the point Q in the ratio 1: 2. Does the line x – 2y = 0 contain Q?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 11

Question 11.
Find the point of intersection of the lines:
4x + 3y = 1 and 3x – y + 9 = 0. If this point lies on the line (2k – 1)x – 2y = 4; find the value of k.
Solution:
Consider the given equations:
4x + 3y = 1 ….(1)
3x – y + 9 = 0 ….(2)
Multiplying (2) with 3, we have:
9x – 3y = -27 ….(3)
Adding (1) and (3), we get,
13x = -26
x = -2
From (2), y = 3x + 9 = -6 + 9 = 3
Thus, the point of intersection of the given lines (1) and (2) is (-2, 3).
The point (-2, 3) lies on the line (2k – 1)x – 2y = 4.
(2k – 1)(-2) – 2(3) = 4
-4k + 2 – 6 = 4
-4k = 8
k = -2

Question 12.
Show that the lines 2x + 5y = 1, x – 3y = 6 and x + 5y + 2 = 0 are concurrent.
Solution:
We know that two or more lines are said to be concurrent if they intersect at a single point.
We first find the point of intersection of the first two lines.
2x + 5y = 1 ….(1)
x – 3y = 6 ….(2)
Multiplying (2) by 2, we get,
2x – 6y = 12 ….(3)
Subtracting (3) from (1), we get,
11y = -11
y = -1
From (2), x = 6 + 3y = 6 – 3 = 3
So, the point of intersection of the first two lines is (3, -1).
If this point lie on the third line, i.e., x + 5y + 2 = 0, then the given lines will be concurrent.
Substituting x = 3 and y = -1, we have:
L.H.S. = x + 5y + 2 = 3 + 5(-1) + 2 = 5 – 5 = 0 = R.H.S.
Thus, (3, -1) also lie on the third line.
Hence, the given lines are concurrent.

Equation of a Line Exercise 14B – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 12
Solution:

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 14
Solution:

Question 3.
Find the slope of the line passing through the following pairs of points:
(i) (-2, -3) and (1, 2)
(ii) (-4, 0) and origin
(iii) (a, -b) and (b, -a)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 16

Question 4.
Find the slope of the line parallel to AB if:
(i) A = (-2, 4) and B = (0, 6)
(ii) A = (0, -3) and B = (-2, 5)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 17

Question 5.
Find the slope of the line perpendicular to AB if:
(i) A = (0, -5) and B = (-2, 4)
(ii) A = (3, -2) and B = (-1, 2)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 18

Question 6.
The line passing through (0, 2) and (-3, -1) is parallel to the line passing through (-1, 5) and (4, a). Find a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 19

Question 7.
The line passing through (-4, -2) and (2, -3) is perpendicular to the line passing through (a, 5) and (2, -1). Find a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 20

Question 8.
Without using the distance formula, show that the points A (4, -2), B (-4, 4) and C (10, 6) are the vertices of a right-angled triangle.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 21

Question 9.
Without using the distance formula, show that the points A (4, 5), B (1, 2), C (4, 3) and D (7, 6) are the vertices of a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 22

Question 10.
(-2, 4), (4, 8), (10, 7) and (11, -5) are the vertices of a quadrilateral. Show that the quadrilateral, obtained on joining the mid-points of its sides, is a parallelogram.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 23

Question 11.
Show that the points P (a, b + c), Q (b, c + a) and R (c, a + b) are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 25

Question 12.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 26
Solution:

Question 13.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 28
Solution:

Question 14.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 30
Solution:

Question 15.
A (5, 4), B (-3, -2) and C (1, -8) are the vertices of a triangle ABC. Find:
(i) the slope of the altitude of AB,
(ii) the slope of the median AD, and
(iii) the slope of the line parallel to AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 32

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 33
Solution:

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 35
Solution:

Question 18.
The points (-3, 2), (2, -1) and (a, 4) are collinear. Find a.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 37

Question 19.
The points (K, 3), (2, -4) and (-K + 1, -2) are collinear. Find K.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 38

Question 20.
Plot the points A (1, 1), B (4, 7) and C (4, 10) on a graph paper. Connect A and B, and also A and C.
Which segment appears to have the steeper slope, AB or AC?
Justify your conclusion by calculating the slopes of AB and AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 39

Question 21.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 40
Solution:

Equation of a Line Exercise 14C – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the equation of a line whose:
y-intercept = 2 and slope = 3.
Solution:
Given, y-intercept = c = 2 and slope = m = 3.
Substituting the values of c and m in the equation y = mx + c, we get,
y = 3x + 2, which is the required equation.

Question 2.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 42
Solution:

Question 3.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 45

Question 4.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 46

Question 5.
Find the equation of the line passing through:
(i) (0, 1) and (1, 2) (ii) (-1, -4) and (3, 0)
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 47

Question 6.
The co-ordinates of two points P and Q are (2, 6) and (-3, 5) respectively. Find:
(i) the gradient of PQ;
(ii) the equation of PQ;
(iii) the co-ordinates of the point where PQ intersects the x-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 48

Question 7.
The co-ordinates of two points A and B are (-3, 4) and (2, -1). Find:
(i) the equation of AB;
(ii) the co-ordinates of the point where the line AB intersects the y-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 49

Question 8.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 50
Solution:

Question 9.
In ΔABC, A = (3, 5), B = (7, 8) and C = (1, -10). Find the equation of the median through A.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 52

Question 10.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 53
Solution:

Question 11.
Find the equation of the straight line passing through origin and the point of intersection of the lines x + 2y = 7 and x – y = 4.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 55

Question 12.
In triangle ABC, the co-ordinates of vertices A, B and C are (4, 7), (-2, 3) and (0, 1) respectively. Find the equation of median through vertex A.
Also, find the equation of the line through vertex B and parallel to AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 56

Question 13.
A, B and C have co-ordinates (0, 3), (4, 4) and (8, 0) respectively. Find the equation of the line through A and perpendicular to BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 57

Question 14.
Find the equation of the perpendicular dropped from the point (-1, 2) onto the line joining the points (1, 4) and (2, 3).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 58

Question 15.
Find the equation of the line, whose:
(i) x-intercept = 5 and y-intercept = 3
(ii) x-intercept = -4 and y-intercept = 6
(iii) x-intercept = -8 and y-intercept = -4
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 59

Question 16.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 60

Question 17.
Find the equation of the line with x-intercept 5 and a point on it (-3, 2).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 61

Question 18.
Find the equation of the line through (1, 3) and making an intercept of 5 on the y-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 62

Question 19.
Find the equations of the lines passing through point (-2, 0) and equally inclined to the co-ordinate axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 63

Question 20.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 64
Solution:

Question 21.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 66
Solution:

Question 22.
A (1, 4), B (3, 2) and C (7, 5) are vertices of a triangle ABC, Find:
(i) the co-ordinates of the centroid of triangle ABC.
(ii) the equation of a line, through the centroid and parallel to AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 68

Question 23.
A (7, -1), B (4, 1) and C (-3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP: CP = 2: 3.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 69

Equation of a Line Exercise 14D – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Find the slope and y-intercept of the line:
(i) y = 4
(ii) ax – by = 0
(iii) 3x – 4y = 5
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 70

Question 2.
The equation of a line x – y = 4. Find its slope and y-intercept. Also, find its inclination.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 71

Question 3.
(i) Is the line 3x + 4y + 7 = 0 perpendicular to the line 28x – 21y + 50 = 0?
(ii) Is the line x – 3y = 4 perpendicular to the line 3x – y = 7?
(iii) Is the line 3x + 2y = 5 parallel to the line x + 2y = 1?
(iv) Determine x so that the slope of the line through (1, 4) and (x, 2) is 2.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 72

Question 4.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 74
Solution:

Question 5.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 76
Solution:

Question 6.
(i) Lines 2x – by + 3 = 0 and ax + 3y = 2 are parallel to each other. Find the relation connecting a and b.
(ii) Lines mx + 3y + 7 = 0 and 5x – ny – 3 = 0 are perpendicular to each other. Find the relation connecting m and n.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 78

Question 7.
Find the value of p if the lines, whose equations are 2x – y + 5 = 0 and px + 3y = 4 are perpendicular to each other.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 79

Question 8.
The equation of a line AB is 2x – 2y + 3 = 0.
(i) Find the slope of the line AB.
(ii) Calculate the angle that the line AB makes with the positive direction of the x-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 80

Question 9.
The lines represented by 4x + 3y = 9 and px – 6y + 3 = 0 are parallel. Find the value of p.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 81

Question 10.
If the lines y = 3x + 7 and 2y + px = 3 are perpendicular to each other, find the value of p.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 82

Question 11.
The line through A(-2,3) and B(4,b) is perpendicular to the line 2x – 4y =5. Find the value of b.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 83

Question 12.
Find the equation of the line through (-5, 7) and parallel to:
(i) x-axis (ii) y-axis
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 84

Question 13.
(i) Find the equation of the line passing through (5, -3) and parallel to x – 3y = 4.
(ii) Find the equation of the line parallel to the line 3x + 2y = 8 and passing through the point (0, 1).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 85

Question 14.
Find the equation of the line passing through (-2, 1) and perpendicular to 4x + 5y = 6.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 86

Question 15.
Find the equation of the perpendicular bisector of the line segment obtained on joining the points (6, -3) and (0, 3).
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 87

Question 16.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 88
Solution:

Question 17.
B (-5, 6) and D (1, 4) are the vertices of rhombus ABCD. Find the equation of diagonal BD and of diagonal AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 90

Question 18.
A = (7, -2) and C = (-1, -6) are the vertices of square ABCD. Find the equations of diagonal BD and of diagonal AC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 91

Question 19.
A (1, -5), B (2, 2) and C (-2, 4) are the vertices of triangle ABC, find the equation of:
(i) the median of the triangle through A.
(ii) the altitude of the triangle through B.
(iii) the line through C and parallel to AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 92

Question 20.
(i) Write down the equation of the line AB, through (3, 2) and perpendicular to the line 2y = 3x + 5.
(ii) AB meets the x-axis at A and the y-axis at B. Write down the co-ordinates of A and B. Calculate the area of triangle OAB, where O is the origin.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 93

Question 21.
The line 4x – 3y + 12 = 0 meets the x-axis at A. Write the co-ordinates of A.
Determine the equation of the line through A and perpendicular to 4x – 3y + 12 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 94

Question 22.
The point P is the foot of perpendicular from A (-5, 7) to the line whose equation is 2x – 3y + 18 = 0. Determine:
(i) the equation of the line AP
(ii) the co-ordinates of P
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 95

Question 23.
The points A, B and C are (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC.
If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 96

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 97
Solution:

Question 25.
Find the value of a for which the points A(a, 3), B(2, 1) and C(5, a) are collinear. Hence, find the equation of the line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 99

Equation of a Line Exercise 14E – Selina Concise Mathematics Class 10 ICSE Solutions

Question 1.
Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5. Find its co-ordinates of point P.
Also, find the equation of the line through P and parallel to 3x + 5y = 7.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 100

Question 2.
The line segment joining the points A(3, -4) and B (-2, 1) is divided in the ratio 1: 3 at point P in it. Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y + 4 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 101

Question 3.
A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.
Solution:

Question 4.
Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.
Solution:

Question 5.
Solution:

Question 6.
(1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
Solution:

Question 7.
Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
Solution:

Question 8.
A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.
Solution:

Question 9.
A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.
Solution:

Question 10.
Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x – 2y = 1.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 111

Question 11.

Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 113

Question 12.
O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:
(i) the equation of median of triangle OAB through vertex O.
(ii) the equation of altitude of triangle OAB through vertex B.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 114

Question 13.
Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.
Does the line 3x = y + 1 bisect the line segment joining the two given points?
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 115

Question 14.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 116

Question 15.
Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 117

Question 16.
The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find:
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point, where the perpendicular through A, as obtained in (i), meets BC.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 118

Question 17.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 119
Solution:

Question 18.
P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 121

Question 19.
A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 122

Question 20.
A line AB meets the x-axis at point A and y-axis at point B. The point P (-4, -2) divides the line segment AB internally such that AP : PB = 1 : 2. Find:
(i) the co-ordinates of A and B.
(ii) the equation of line through P and perpendicular to AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 123

Question 21.
A line intersects x-axis at point (-2, 0) and cuts off an intercept of 3 units from the positive side of y-axis. Find the equation of the line.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 124

Question 22.
Find the equation of a line passing through the point (2, 3) and having the x-intercept of 4 units.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 125

Question 23.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 126
Solution:

Question 24.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 146
Solution:

Question 25.
The ordinate of a point lying on the line joining the points (6, 4) and (7, -5) is -23. Find the co-ordinates of that point.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 157
Question 26.
Points A and B have coordinates (7, -3) and (1, 9) respectively. Find:
(i) the slope of AB.
(ii) the equation of the perpendicular bisector of the line segment AB.
(iii) the value of ‘p’ if (-2, p) lies on it.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 149

Question 27.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 150
Solution:

Question 28.
The equation of a line 3x + 4y – 7 = 0. Find:
(i) the slope of the line.
(ii) the equation of a line perpendicular to the given line and passing through the intersection of the lines x – y + 2 = 0 and 3x + y – 10 = 0.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 151
Question 29.
ABCD is a parallelogram where A(x, y), B(5, 8), C(4, 7) and D(2, -4). Find:
(i) Co-ordinates of A
(ii) Equation of diagonal BD
Solution:

Question 30.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 153
Solution:

Question 31.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 156
Solution:

Question 32.
Find the equation of the line that has x-intercept = -3 and is perpendicular to 3x + 5y = 1.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 129

Question 33.
A straight line passes through the points P(-1, 4) and Q(5, -2). It intersects x-axis at point A and y-axis at point B. M is the mid- t point of the line segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of points A and B.
(iii) the co-ordinates of point M
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 130

Question 34.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 131
Solution:

Question 35.
A line through point P(4, 3) meets x-axis at point A and the y-axis at point B. If BP is double of PA, find the equation of AB.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 133

Question 36.
Find the equation of line through the intersection of lines 2x – y = 1 and 3x + 2y = -9 and making an angle of 30° with positive direction of x-axis.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 134

Question 37.
Find the equation of the line through the Points A(-1, 3) and B(0, 2). Hence, show that the points A, B and C(1, 1) are collinear.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 135

Question 38.
Three vertices of a parallelogram ABCD taken in order are A(3, 6), B(5, 10) and C(3, 2), find :
(i) the co-ordinates of the fourth vertex D.
(ii) length of diagonal BD.
(iii) equation of side AB of the parallelogram ABCD.
Solution:
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 136

Question 39.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 137
Solution:

Question 40.
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 139

Question 41.

i. Since A lies on the X-axis, let the co-ordinates of A be (x, 0).
Since B lies on the Y-axis, let the co-ordinates of B be (0, y).
Let m = 1 and n = 2
Using Section formula,
Selina Concise Mathematics Class 10 ICSE Solutions Equation of a Line - 140
⇒ Slope of line perpendicular to AB = m = -2
P = (4, -1)
Thus, the required equation is
y – y₁ = m(x – x₁)
⇒ y – (-1) = -2(x – 4)
⇒ y + 1 = -2x + 8
⇒ 2x + y = 7

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