Selina Concise Mathematics Class 7 ICSE Solutions Chapter 13 Set Concepts

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 13 Set Concepts (Some Simple Divisions by Vedic Method)

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 13 Set Concepts (Some Simple Divisions by Vedic Method)

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POINTS TO REMEMBER
1. Definition of a Set : In our day to day life, different collective nouns are used to describe collection of objects ; such as : a group of students playing cricket, a pack of cards, a bunch of flowers, etc
In mathematics, such collections of objects are named as sets.
A set is a collection of well-defined objects, things or symbols, etc.
The phrase ‘well-defined’ means ; it must be possible to know, without any doubt, whether a given object (thing or symbol) belongs to the set under consideration or not.
For example :
“The set of tall boys of Class 10” is not well-defined ; since it is not possible to know that which boys are to be included and exactly what is the limit.
But when we say, “The set of boys of Class 10, which are taller than Peter”, now we can compare the heights of different boys with the height of Peter and can know exactly, that which boys are to be included in the required set. Thus, the objects are well-defined.
2. Elements of a set: The objects (things, symbols, etc.) used to form a set are called elements or members of the set.
In general, a set is denoted by a capital letter of English alphabet with its elements written inside curly braces and separated by commas.
e.g., Set A = {5, 10, 12, 15}
3. Use of Symbol ‘∈’ or Symbol ‘∉ ‘ : The symbol ‘∈’ stands for ‘belongs to’ or ‘ is an element of’ or ‘is a member of’; whereas the symbol ‘∉’ stands for ‘does not belong to’ or ‘is not an element of or ‘is not a member of’.
e.g., For set P = {3, 6, 8, 13, 18} ; 3 ∈ P, 5∉ P and so on.
(i) The elements in a set can be written in any order.
Thus, {a, b, c, d] is the same set as {b, d, a, c} or {c, b, d, a}, etc.
(ii) The elements in a set should not be repeated, i.e. if any element occurs many times, it should be written only once.
Thus, set of letters of the word ‘crook’ = {c, r, o, k}.
There are two os in the given word “crook” ; but in the set, it is written only once.
4. Representation of A Set: A set, in general, is represented in :
(i) Description method (form)
(ii) Tabular or Roster method (form)
(iii) Set-builder or Rule method
For example :
N is the set of natural numbers [Description method]
N = {1, 2, 3, 4, 5, ……. } [Roster or Tabular method]
N = {x : x is a natural number}, or {x : x ∈ N} [Set-builder or Rule method]
[The symbol ‘ : ’ stands for such that and the set {x : x ∈ N} is read as, “the set of x such that x is a natural number”].
It is clear from the example given above that:
(i) in description method a well-defined description about the set is given.
(ii) in roster or tabular method the elements of the set are written inside a pair of curly braces and are separated by commas.
(iii) in set-builder or ruler method the actual elements of the set are not written, but a rule or a statement or a formula is written in the briefest possible way.
5. Cardinal Number : The cardinal number of a set is the number of elements in it.
Thus, if a set A has 5 elements ; its cardinal number is 5 and we represent it by writing n (A) = 5 .
Similarly, if set B = Set of even natural numbers less than 10 then, B = {2, 4, 6, 8} and n (B) = 4.
If B = {0}, then n (B) = 1. Since, 0 is an element of set B.
6. Types of Sets :
(i) Finite Set: A set is said to be a finite set, if it has a limited (countable) number of elements in it.
For example :
(a) S = Set of natural numbers between 10 and 15 = {11, 12, 13, 14}
(b) P = {0, 1,2, , 20} = {x : x ∈ W and x ≤ 20} and so on.
(ii) Infinite set: A set is said to be an infinite set, if it has an unlimited (uncountable) number of elements in it.
For example :
(a) P = Set of prime numbers = {2, 3, 5,…. }
(b) B = (x : x ∈ N and x ≥ 21} = (21, 22, 23, } ans so on.
(iii) Empty set or Null set: The set, with no element in it, is called the empty set or the null set.
The empty set is represented by a pair of braces with no element in it or by the Danish letter Φ, which is pronounced as ‘oe’
Thus, the empty set = { } = Φ
Note : For empty set, it is wrong to call ‘an empty set’ or ‘a null set’ as there is one and only one empty set though it may have many descriptions.
Therefore, it is always called “the empty set or the null set”.
Some examples of the empty set :
(a) Let A = {a man of age more than 400 years}.
Since there can not be any man with the age more than 400 years; the set A will have no element in it i.e. It is the empty set. And we write : A = { } or Φ
(b) If B = (Triangles with 4 sides} ; it is clear that B = Φ
Note :
1. Φ ≠{0}, since {0} is a set with 0 as its element whereas Φ has no element.
2. {Φ} ≠ {0}, since both the sets have different elements.
3. The cardinal number of the empty set is 0 i.e. n (Φ) = 0.
(iv) Disjoint sets : Sets having no element in common are called disjoint sets.
For example :
Sets P = (5, 7, 9} and O = (4, 6, 10, 12} are disjoint; as they do not have any element in common.
(v) Joint (overlapping) sets : Sets having atleast one element in common are called joint or overlapping sets.
For example:
Set B = (4,6, 8,10,12} and set C = (3,6,9,12,15} are joint sets; as they have elements 6 and 12 common.
(vi) Equal sets : Two sets are said to be equal, if the elements of both the sets are the same.
For example :
If set A = {x, y, z} and set B = {last three letters of English alphabet}.
Clearly, sets A and B have the same elements and so set A = set B.
(vii) Equivalent sets : Two sets are said to be equivalent, if they have equal number of elements in them, i.e. the cardinal numbers of both the sets are equal.
For example :
Let A = (3, 6, 9} and B = {a, b, c}.
Since, set A has 3 elements and set B also has 3 elements i.e., n (A) = n (B); therefore, sets A and B are equivalent and for this, we write : A ↔ B.
Note:
1. Equal sets are always equivalent; but the converse is not always true (i.e. it is not necessary that equivalent sets are equal also).
2. In equivalent sets, the number of elements (cardinal number) are equal, whereas in equal sets the element are the same.
3. Two infinite sets are always equivalent;

Set Concepts Exercise 13A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Find, whether or not, each of the following collections represent a set:
(i) The collection of good students in your school.
(ii) The collection of the numbers between 30 and 45.
(iii) The collection of fat-people in your colony.
(iv) The collection of interesting books in your school library.
(v) The collection of books in the library and are of your interest.
Solution:
(i) It is not a set as it is not well defined.
(ii) It is a set
(iii) It is not a set as it is not well defined.
(iv) It is not a set as it is not well defined.
(v) It is a set.

Question 2.
State whether true or false :
(i) Set {4, 5, 8} is same as the set {5, 4, 8} and the set {8, 4, 5}
(ii) Sets {a, b, m, n} and {a, a, m, b, n, n) are same.
(iii) Set of letters in the word ‘suchismita’ is {s, u, c, h, i, m, t, a}
(iv) Set of letters in the word ‘MAHMOOD’ is {M, A, H, O, D}.
Solution:
(i) True
(ii) True
(iii) True as it has the same elements
(iv) True as it has the same elements.

Question 3.
Let set A = {6, 8, 10, 12} and set B = {3, 9, 15, 18}.
Insert the symbol ‘ ∈ ’ or ‘ ∉ ’ to make each of the following true :
(i) 6 …. A
(ii) 10 …. B
(iii) 18 …. B
(iv) (6 + 3) …. B
(v) (15 – 9) …. B
(vi) 12 …. A
(vii) (6 + 8) …. A
(viii) 6 and 8 …. A
Solution:
(i) 6 ∈ A
(ii) 10 ∉ B
(iii) 18 ∈B
(iv) (6 + 3) or 9 ∈ B
(v) 15 – 9 or 6 g B
(vi) 12 ∈ A
(vii) 6 + 8 or 14 ∉ A
(viii) 6 and 8 ∈ A

Question 4.
Express each of the following sets in
roster form :
(i) Set of odd whole numbers between 15 and 27.
(ii) A = Set of letters in the word “CHITAMBARAM”
(iii) B = {All even numbers from 15 to 26}
(iv) P = {x : x is a vowel used in the word ‘ARITHMETIC’}
(v) S = {Squares of first eight whole numbers}
(vi) Set of all integers between 7 and 94; which are divisible by 6.
(vii) C = {All composite numbers between 2 and 20}
(viii) D = Set of Prime numbers from 2 to 23.
(ix) E = Set of natural numbers below 30 which are divisible by 2 or 5.
(x) F = Set of factors of 24.
(xi) G = Set of names of three closed figures in Geometry.
(xii) H = {x : x eW and x < 10}
(xiii) J = {x: x e N and 2x – 3 ≤17}
(xiv) K = {x : x is an integer and – 3 < x < 5}
Solution:
(i) {17, 19, 21, 23, 25}
(ii) A = (C, H, I, T, A, M, B, R}
(iii) B = {16, 18, 20, 22, 24, 26}
(iv) P = {a, e, i}
(v) S = {0, 1, 4, 9, 16, 25, 36, 49}
(vi) {12, 18, 24; 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90}
(vii) C = {4, 6, 8, 9, 10, 12, 14, 15, 16, 18}
(viii) D = {2, 3, 5, 7, 11, 13, 17, 19,23}
(ix) E = {2, 4, 5, 6, 8, 10, 12, 14, 15, 16, 18, 20, 22, 24, 25, 26, 28}
(x) F={l,2, 3, 4, 6, 8, 12, 24}
(xi) G = {Triangle, quadrilateral, circle}
(xii) H = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
(xiii) 2x – 3 ≤ 17
⇒ 2x ≤ 17 + 3 2 x ≤ 20
⇒ x ≤ \(\frac { 20 }{ 2 }\)
x ≤ 10
∴ J = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
(xiv) ∵ – 3 < x < 5
∴x lies between – 3 and 5
∴K = {- 2, – 1, 0, 1, 2, 3, 4}

Question 5.
Express each of the following sets in set- builder notation (form) :
(i) {3, 6, 9, 12, 15}
(ii) {2, 3, 5, 7, 11, 13 …. }
(iii) {1, 4, 9, 16, 25, 36}
(iv) {0, 2, 4, 6, 8, 10, 12, …. }
(v) {Monday, Tuesday, Wednesday}
(vi) {23, 25, 27, 29, … }
(vii) {\(\frac { 1 }{ 3 }\),\(\frac { 1 }{ 4 }\),\(\frac { 1 }{ 5 }\),\(\frac { 1 }{ 6 }\),\(\frac { 1 }{ 7 }\),\(\frac { 1 }{ 8 }\)}
(viii) {42, 49, 56, 63, 70, 77}
Solution:
(i) {3, 6, 9, 12, 15}
= {x: x is a natural number divisible by 3 ;x< 18}
(ii) {2, 3, 5, 7, 11, 13, }
= {x : x is a prime number}
(iii) {1,4,9, 16,25,36}
= {x : x is a perfect square ; x < 36}
(iv) {0, 2, 4, 6, 8, 10, 12, }
= {x : x is a whole number divisible by 2}
(v) {Monday, Tuesday, Wednesday}
= {x : x is one of the first three days of 3 week}
(vi) {23, 25, 27, 29, }
= {x : x is an odd natural number; x ≥ 23}
(vii) {\(\frac { 1 }{ 3 }\),\(\frac { 1 }{ 4 }\),\(\frac { 1 }{ 5 }\),\(\frac { 1 }{ 6 }\),\(\frac { 1 }{ 7 }\),\(\frac { 1 }{ 8 }\)}
= {x: x = \(\frac { 1 }{ n }\) when n is a natural number: 3 ≤ n ≤ 8}
(viii) {42, 49, 56, 63, 70, 77}
= (x: x is a natural number divisible by 7 ; 42 ≤x ≤ 77}

Question 6.
Given : A = {x : x is a multiple of 2 and is less than 25}
B = {x : x is a square of a natural number and is less than 25}
C = {x : x is a multiple of 3 and is less than 25}
D = {x: x is a prime number less than 25}
Write the sets A, B, C and D in roster form.
Solution:
A = {x: x is a multiple of 2 and is less than 25}
= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24}
B = {x : x is a square of natural number and is less than 25}
= {1,4,9,16}
C = {x : x is a multiple of 3 and is less than 25}
= {3, 6, 9, 12, 15, 18,21,24}
D = {x : x is a prime number less than 25}
= {2, 3, 5, 7, 11, 13, 17, 19, 23}

Set Concepts Exercise 13B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Write the cardinal number of each of the following sets:
(i) A = Set of days in a leap year.
(ii) B = Set of numbers on a clopk-face.
(iii) C = {x : x ∈ N and x ≤ 7}
(iv) D = Set of letters in the word “PANIPAT”.
(v) E = Set of prime numbers between 5 and 15.
(vi) F = {x : x ∈ Z and – 2 < x ≤ 5}
(vii) G = {x : x is a perfect square number, x ∈N and x ≤ 30}.
Solution:
(i) n A = 366
(ii) n B = 12
(iii) n C =7
(iv) n D = 5
(v) n E = 3
(vi) n F =7
(vii) n G = 5

Question 2.
For each set, given below, state whether it is finite set, infinite set or the null set :
(i) {natural numbers more than 100}
(ii) A = {x : x is an integer between 1 and 2}
(iii) B = {x : x ∈ W ; x is less than 100}.
(iv) Set of mountains in the world.
(v) {multiples of 8}.
(vi) {even numbers not divisible by 2}.
(vii) {squares of natural numbers}.
(viii) {coins used in India}
(ix) C = {x | x is a prime number between 7 and 10}.
(x) Planets of the Solar system.
Solution:
(i) {Natural numbers more than 100}
= It is an infinite set
(ii) A = {x : x is an integer between 1 and 2}
It is a null set
(iii) B = {x : x ∈ W, x is less than 100}
It is finite set as it has 100 elements i.e. from 0 to 99.
(iv) Set of mountains in the world.
∴ It is an infinite set
(v) {Multiples of 8}
It is an infinite set
(vi) {Even numbers not divisible by 2}
It is a null set
(vii) {Squares of natural numbers}
∴ It is an infinite set
(viii) {Coins used in India}
∴It is a finite set as these are countable
(ix) {x | x is a prime number between 7 and 10}
As there is not such prime number between 7 and 10.
Hence it is null set
(x) Planets of two Solar system.
It is finite set as there are countable.

Question 3.
State, which of the following pairs of sets are disjoint :
(i) {0, 1, 2, 6, 8} and {odd numbers less than 10.
(ii) {birds} and {tress}
(iii) {x : x is a fan of cricket} and {x : x is a fan of football}.
(iv) A = {natural numbers less than 10} and B = {x : x is a multiple of 5}.
(v) {people living in Calcutta} and {people living in West Bengal}.
Solution:
(i) {0, 1, 2, 6, 8} and {odd numbers less than 10}
⇒ {0, 1,2, 6, 8} and {1,3, 5, 7, 9}
∴There sets are not disjoint sets as there is one element (1) is common.
(ii) {Birds} and {trees}
These are disjoint sets as there is no common element in term
(iii) {x : x is a fan of cricket} and {x : x is a fan of football}
These are not disjoint sets as there can be a person who is fan of both the games.
(iv) A = {Natural numbers less than 10} and B = {x : x is a multiple of 5}
⇒ A = {1, 2, 3, 4, 5, 6, 7, 8, 9} and B = {5, 10, 15 }
These are hot disjoint sets as there is one element 5, which is common.
(v) {People living in Calcutta} and {People living in West Bengal}.
These are not disjoint sets as people of Calcutta are the people of West Bengal as Calcutta is a city of West Bengal.
So, only (ii) is a pair of disjoint sets.

Question 4.
State whether the given pairs of sets are equal or equivalent.
(i) A = {first four natural numbers} and B = {first four whole numbers}.
(ii) A = Set of letters of the word “FOLLOW” and B = Set of letters of the word “WOLF”.
(iii) E = {even natural numbers less than 10} and O = {odd natural numbers less than 9}
(iv) A = {days of the week starting with letter S} and B = {days of the week starting with letter T}.
(v) M = {multiples of 2 and 3 between 10 and 20} and N = {multiples of 2 and 5 between 10 and 20}.
(vi) P = {prime numbers which divide 70 exactly} and Q = {prime numbers which divide 105 exactly}
(vii) A = {0², 1², 2², 3², 4²} and = {16, 9,4, 1, 0}.
(viii) E = {8,JO, 12, 14, 16} and F = {even natural numbers between 6 and 18}.
(ix) A = {letters of the word SUPERSTITION} and B = {letters of the word JURISDICTION}.
Solution:
(i) A = {first four natural numbers}
= {1,2, 3, 4}
B = {first first whole number)
= {0, 1,2,3}
These are equivalent sets as both have equal number of elements but not same.
(ii) A = Set of letters of the word ‘FOLLOW’
= {F, O, L, W}
B = Set of letters of the word ‘WOLF’
= {W, O, L, F}
These are equal sets as these have same and equal elements.
(iii) E = {even natural numbers less than 10}
= {2, 4, 6, 8}
O = {odd natural numbers less than 9}
= {L3, 5, 7}
These are equivalent sets as both have equal number of elements but not the same.
(iv) A = {Days of the week starting with letter S}
= {Sunday, Saturday}
B = {Days of the week starting with letter T}
= {Tuesday, Thursday}
These are equivalent sets as both have equal number of elements.
(v) M = {Multiples of 2 and 3 between 10 and 20}
= {12, 14, 15, 16, 18}
N = {Multiples of 2 and 5 between 10 and 20}
= {12, 14, 15, 16, 18}
These are equal sets as these have same and equal number of elements.
(vi) P = {Prime numbers which divide 70 exactly}
= {2, 5, 7}
Q = {Prime numbers which divide 105 exactly}
= {3, 5, 7}
These are equivalent sets as these have equal number of elements.
(vii) A = {02, l2, 22, 32, 42} = {0, 1, 4, 9, 16} B = {16, 9, 4, 1,0}
These are equal sets as these have same and equal number of elements.
(viii) E = {8, 10, 12, 14, 16}
F = {even natural numbers between 6 and 18}
= {8, 10, 12, 14, 16}
These sets are equal as these have same and equal number of elements
(ix) A = {Letters of the word SUPERSTITION}
= {S, U, P, E, R, T, I, O, N}
B = Letters of the word JURISDICTION.
= (J, U, R, I, S, D, C, T, O, N}
These are neither equal nor equivalent sets as these have different and unequal elements.

Question 5.
Examine which of the following sets are the empty sets :
(i) The set of triangles having three equal sides.
(ii) The set of lions in your class.
(iii) {x 😡 + 3 = 2 and xeN}
(iv) P = {x : 3x = 0}
Solution:
(i) The set of triangle having three equal sides. This is not an empty set
(ii) The set of lions in your class This is an empty set
(iii) {x : x +3 = 2 and x ∈N}
x ≠ 3 = 2 ⇒ x = 2-3= -l
which is not a natural number.
∴ It is an empty set.
(iv) P = {x : 3x = 0} = {0} which is not an empty set.
Hence (ii) and (iii) are empty sets.

Question 6.
State true or false :
(i) All examples of the empty set are equal.
(ii) All examples of the empty set are equivalent.
(iii) If two sets have the same cardinal number, they are equal sets.
(iv) If n (A) = n (B) then A and B are equivalent sets.
(v) If B = {x : x + 4 = 4}, then B is the empty set.
(vi) The set of all points in a line is a finite set.
(vii) The set of letters in your Mathematics book is an infinite set.
(viii) If M = {1, 2, 4, 6} and N = {x : x is a factor of 12} ; then M = N.
(ix) The set of whole numbers greater than 50 is an infinite set.
(x) If A and B are two different infinite sets, then n (A) = n (B).
Solution:
(i) True
(ii) True
(iii) False
(iv) True
(v) False
(vi) False
(vii) False
(viii) False
( ix )True
(x) False

Question 7.
Which of the following represent the null set ?
φ, {0}, 0, { }, {φ}

Solution:
φ and { } are the null sets other are not as there have same element.

Set Concepts Exercise 13C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Fill in the blanks :
(i) If each element of set P is also an element of set Q, then P is said to be …… of Q and Q is said to be of P.
(ii) Every set is a ….. of itself.
(iii) The empty set is a …… of every set.
(iv) If A is proper subset of B, then n (A) …. n (B).
Solution:
(i) If each element of set P is also an element of set Q then P is said to be subject of Q ; and Q is said to be super set of P.
(ii) Every set is a subset of itself.
(iii) The empty set is subset of every set.
(iv) If A is proper subset of B, then n (A) is less than n (B)

Question 2.
If A = {5, 7, 8, 9} ; then which of the following are subsets of A ?
(i) B = {5, 8}
(ii) C = {0}
(iii) D = {7, 9, 10}
(iv) E = { }
(V) F = {8, 7, 9, 5}
Solution:
(i) B = {5, 8,}
∴B ⊂ A
(ii) C = {0}
∴ C φ A
(iii) D = {7, 9, 10}
∴D ⊄ A
(iv) E = { }
∴E ⊂A (An empty set is subset of every set)
(v) F = (8, 7, 9, 5}
∴F ⊂A
∵ Every set is subset of it self.
Hence (i), (iv)
and (v) are subsets of A.

Question 3.
If P = {2, 3, 4, 5} ; then which of the following are proper subsets of P ?
(i) A = {3, 4}
(ii) B = { }
(iii) C = {23, 45}
(iv) D = {6, 5, 4}
(v) E = {0}
Solution:
P = {2, 3, 4, 5}
(i) A = {3,4},
(ii) B = { }, C = {23, 45},
D = {6, 5, 4} and E = {0}.
We see that only A and B are the proper subset of P.

Question 4.
If A = {even numbers less than 12},
B = {2, 4},
C = {1, 2, 3},
D = {2, 6} and E = {4}
State which of the following statements are true :
(i) B⊂A
(ii) C⊆A
(iii) D⊂C
(iv) D ⊄ A
(v)E⊇B
(vi) A⊇B⊇E
Solution:
A = {Even number less than 12} = {2, 4, 6, 8, 10}
B = {2, 4}, C = {1, 2, 3},
D = {2, 6} and E = {4}
(i) B ⊂ A: It is true
(ii) C ⊆ A: It is false
(iii) D ⊂ C : It is false
(iv) D ⊄ A
(v) E ⊇ B : It is false
(vi) A ⊇ B ⊇ E : It is true

Question 5.
Given A = {a, c}, B = {p, q, r} and C = Set of digits used to form number 1351.
Write all the subsets of sets A, B and C.
Solution:
(i) A = {a, c}
∴ Subsets are : { } or φ, {a}, {c} and {a, c}
(ii) B = {p,q, r)
∴ subsets are : { } or φ, {p}, {q}, {r}, {p, q}, ip, r}, {q, r} and {p, q, r}
(iii) C = Set of digits used in 135, = {1,3,5}
∴ Subsets are = { }
or φ, {1}, {3}, {5}, {1,3}, {1,5}, {2,5} and {1, 3, 5}

Question 6.
(i) If A = {p, q, r}, then number of subsets of A = ……
(ii) If B = {5, 4, 6, 8}, then number of proper subsets of B = ……
(iii) If C = {0}, then number of subsets of C = …..
(iv) If M = {x : x ∈ N and x < 3}, then M has …… proper subsets.
Solution:
(i) If A = {p, q, r},
then number of subsets of A = 2³ = 2×2×2 = 8
(ii) If B = {5, 4, 6, 8},
then number of proper subsets of B = 24 – 1 = 2 × 2 × 2× 2 – 1 = 16 – 1 = 15
(iii) If C = {0},
then number of subsets of C = 21 = 2
(iv) If M = {x: x ∈ N and x < 3}, = {1, 2}
Then M has proper subsets = 22—1 = 4 — 1 = 3

Question 7.
For the universal set {4, 5, 6, 7, 8, 9, 10, 11,12,13} ; find its subsets A, B, C and D such that
(i) A = {even numbers}
(ii) B = {odd numbers greater than 8}
(iii) C = {prime numbers}
(iv) D = {even numbers less than 10}.
Also, find compliments of each set i.e., find A’, B’, C’ and D’.
Solution:
(i) A = {even numbers}
= {4, 6, 8, 10, 12}
(ii) B = {odd numbers greater then 8}
= {9, 11, 13}
(iii) C = {Prime numbers}
= {5, 7, 11, 13}
(iv) D = {even numbers less than 10}
= {4, 6, 8}
A’= {5, 7, 9, 11, 13},
B’ = {4, 5, 6, 7, 8, 10, 12}
C’ = { 4, 6, 8, 9, 10, 12}
and D’ = {5, 7, 9, 10, 11, 12, 13}

Set Concepts Exercise 13D – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
If A = {4, 5, 6, 7, 8} and B = {6, 8, 10, 12}, find :
(i) A∪B
(ii) A∩B
(iii) A-B
(iv) B-A
Solution:
(i) A∪B
= [All the elements from set A and all the elements from set B]
= {4, 5, 6, 7, 8, 10, 12}
(ii) A∩B
= [elements common to both the sets A and B]
= {6, 8}
(iii) A-B
= [elements of set A which are not in set B]
= {4, 5, 7}
(iv) B-A
= [elements of set B which are not in set A]
= {10, 12}

Question 2.
If A = {3, 5, 7, 9, 11} and B = {4, 7, 10}, find:
(i) n(A)
(ii) n(B)
(iii) A∪B and n(A∪B)
(iv) A∩B and n(A∩B)
Solution:
(i) n(A) = (3, 5, 7, 9, 11) = 5
(ii) n(B) = (4, 7, 10) = 3
(iii) A ∪ B = {3, 4, 5, 7, 9, 10, 11} n(A ∪ B) = 7
(iv) A∩B = {6} n(A∩B)=l

Question 3.
If A = {2, 4, 6, 8} and B = {3, 6, 9, 12}, find:
(i) (A ∩ B) and n(A ∩ B)
(ii) (A – B) and n(A – B)
(iii) n(B)
Solution:
(i) (A ∩ B) = {2, 4, 8} n( A ∩ B) = 3
(ii) (A – B) and n(A – B)
⇒ (A – B) = (2, 4, 8)
⇒ n(A-B) = 3
(iii) n(B) = {3, 6, 9, 12} = 4

Question 4.
If P = {x : x is a factor of 12} and Q = {x: x is a factor of 16}, find :
(i) n(P)
(ii) n(Q)
(iii) Q – P and n(Q – P)
Solution:
(i) n(P) = Factors of 12 are
= 1, 2, 3, 4, 6, 12
∴ n(P) = 6
(ii) n(Q) = Factors of 16 are = 1. 2, 4, 8, 16
∴n(Q) = 5
(iii) Q – P and n(Q – P)
Elements of set P = {1, 2, 3, 4, 6, 12}
Elements of set Q = {1, 2, 4, 8, 16}
∴ Q – P = 8, 16
n(Q-P) = 2

Question 5.
M = {x : x is a natural number between 0 and 8) and N = {x : x is a natural number from 5 to 10}. Find :
(i) M – N and n(M – N)
(ii) N – M and n(N – M)
Solution:
Natural numbers between 0 and 8 M = {0, 1, 2, 3, 4, 5, 6, 7} and Natural numbers between 5 to 10 N = {6, 7, 8, 9, 10}
(i) M – N = {1, 2, 3, 4} and n(M – N) = 4
(ii) N – M = {8, 9, 10} and n (N – M) = 3

Question 6.
If A = {x: x is natural number divisible by 2 and x< 16} and B = {x:x is a whole number divisible by 3 and x < 18}, find :
(i) n(A)
(ii) n(B)
(iii) A∩B and n(A∩B)
(iv) n(A – B)
Solution:
(i) A = {x : x is natural number divisible-by 2 and x < 16}
A = {2, 4, 6, 8, 10, 12, 14}
n(A) = 7
(ii) B = {x: x is a whole number divisible by 3 and x < 18}
B = {3, 6, 9, 12, 15, 18}
n(B) = 6
(iii) A n B = {2, 4, 6, 8, 10, 12, 14} n {3, 6, 9, 12, 15, 18}
A∩B = {6,12} n(A ∩ B) = 2
(iv) A – B = {2, 4, 6, 8, 10, 12, 14} – {3, 6, 9, 12, 15, 18}
A-B = {2,4, 8, 10, 14} n(A – B) = 5

Question 7.
Let A and B be two sets such that n(A) = 75, M(B) = 65 and n(A ∩ B) = 45, find :
(i) n(A∪ B)
(ii) n(A – B)
(iii) n(B – A)
Solution:
n(A ∩ B)
n(A) = 75, n(B) = 65 and n(A ∩ B) = 45
(i) We know that,
n( A ∪B) = n(A) + n(B) – n( A ∩ B)
n(A ∪B) =75 + 65 – 45
n(A∪B) = 140-45 = 95
(ii) We know that,
n(A – B) = n(A) – n(A ∩ B)
n(A – B) = 75 – 45 = 30
(iii) We know that,
n(B – A) = n(B) – n(A ∩ B)
n(B – A) = 65 – 45 = 20

Question 8.
Let A and B be two sets such that n(A) = 45, n(B) = 38 and n(A ∪B) = 70, find :
(i) n(A∩B)
(ii) n(A-B)
(iii) n(B – A)
Solution:
n(A) = 45, n(B) = 38 and n(A∪ B) = 70
(i) We know that,
n(A ∩ B) = n(A) + M(B) – n(A ∪B)
n(A ∩ B) = 45 + 38 – 70 = 83 – 70 = 13
(ii) We know that,
n(A-B) = n(A ∪B)-n(B)
n(A – B) = 70 – 38 = 32
(iii) We know that,
n(B – A) = n(A ∪ B) – n(A)
n(B – A) = 70 – 45 = 25

Question 9.
Let n(A) 30, n(B) = 27 and n(A∪B) = 45, find :
(i) n(A∩B)
(ii) n(A-B)
Solution:
n(A) = 30, n(B) = 27 and n(A ∪ B) = 45
(i) We know that,
n(A ∩ B) = n( A) + n(B) – n( A∪ B)
n(A ∩ B) = 30 + 27 – 45
n(A ∩ B) = 57 – 45 = 12
(ii) We know that,
n(A-B) = n(A ∪B) – n(B)
n(A – B) = 45 – 27 = 18

Question 10.
Let n(A) = 31, n(B) = 20 and n(A ∩ B) = 6, find:
(i) n(A-B)
(ii) n(B – A)
(iii) n(A ∪B)
Solution:
n(A) = 31, n(B) = 20 and n(A ∩ B) = 6
(i) We know that,
n(A – B) = n(A) – n(A ∩ B)
n(A -B) = 31 -6 = 25
(ii) We know that,
n(B – A) = n(B) – n(A n B)
n(B – A) = 20 – 6 = 14
(iii) We know that,
n(A ∪B) = n(A) + n(B) – n(A ∪ B) n(A∪B) = 31 +20-6 = 45

 

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry (Including Reflection and Rotation)

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 17 Symmetry (Including Reflection and Rotation)

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POINTS TO REMEMBER
1. Symmetry : A geometrical figure is said to be symmetric about a line if on folding about that line, the two parts of the figure exactly concide each other. The given figure is symmetric about the line PQ. The line is said to be a line of symmetry or an axis of symmetry
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 1

2. Lines of symmetry of given geometrical figures :
It is not necessary that every figure under consideration will definitely have a line symmetry. If we consider different types of triangle ; we find :
1. A scalene triangle has no line of Symmetry : i.e. we can not have a line in a scalene triangle about which if the figure (triangle) is folded, the two parts of the figure will coincide.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 2
2. An isosceles triangle has only one line of symmetry. The bisector of angle of vertex which is also the perpendicular bisector of its base.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 3
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 4

The bisectors of the angle of vertices which are also the perpendicular bisectors of its sides.
4. Line/lincs of symmetry of differed types of quadrilaterals are shown below by dotted lines :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 5
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 6

5. In each of the following, the dotted line/lines are the line/lines of symmetry of the given figure:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 7

6. As shown below:
(i) a circle has infinite lines of symmetry ; every line through its centre is line of symmetry’.
(ii) a semi-circle has one line of symmetry.
(iii) a quadrant (one-fourth) of a circle has one line of symmetry’ and so on.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 8
Note : It is clear from the question numbers 4 and 5, given above that:
1. The largest number of lines of symmetry’ of a triangle is three (3).
2. The largest number of lines of symmetry of a quadrilateral is four (4).
i. e. as the number of sides in a triangle is 3 ; the largest number of lines of symmetry in it is 3 and as the number of sides in a quadrilateral is 4 ; the largest number of lines of symmetry is 4.
In the same way :
1. The largest number of lines of symmetry of a pentagon is 5, as a pentagon has 5 sides.
2. The hexagon has 6 sides and so the largest number of lines of symmetry’ of a hexagon is 6.
In general, we can say, that if a polygon has n sides ; M
the largest number of lines of symmetry, it can have, is n.
3. Reflection (Image): Tire given figure shows a candle place ‘d’ distance before a plane mirror MM’, the image of the candle is obtained in the mirror at the same distance ‘d’ behind the mirror. Geometrically, the line joining the candle (c) and its reflection c’ is (Candle) perpendicular bisector of the mirror line MM’.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 9
4. Reflection in x-axis : Reflection if x-axis means the x-axis is considered as the plane mirror, the given point as the object and then to find its image.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 10
Let P (x, y) be a point and. as shown in the figure, when it is reflected in x-axis to point P’; the co-ordinates of image point P’ are (x, – y).
i. e. reflection of P (x, y) in x-axis = P’ (x, – y)
In other words :
Image of P (x, y) in x-axis = P’ (x, – y)
We can say, when a point (x, y) is reflected in x-axis, the sign of its second component (ordinate) changes i.e. tire sign of y changes and so the image of (x, y) in x-axis is (x, – y).
5. Reflection in y-axis : As is clear from the figure, given alongside, the reflection P (x, y) in y-axis is point P’ (- x, y).
We can say, when a point (x,y) is reflected in y-axis, the sign of its first component (abscissa) change i.e. the sign of x changes and so the image of (x, y) in y- axis is (-x,y).
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 11
6. Reflection in Origin : When a point P (x, y) is reflected in origin, the sign of both of its components change i.e. the image of P (x, y) is P’ (- x, -y) as shown along side in the figure.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 12

Symmetry Exercise 17A – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
For each figure, given below, draw the line (s) of symmetry, if possible :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 13
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 14

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 15

Question 2.
Write capital letters A to Z of English alphabet ; and in each case, if possible, draw the largest number of lines of symmetry.
Solution:
Line or lines of symmetry’ is possible in the following alphabets. For others alphabets it is not possible
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 16
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 17

Question 3.
By drawing a free hand sketch of each of the following, draw in each case, the line (s) of symmetry,
if any:
(i) a scalene triangle
(ii) an isosceles right angled triangle
(iii) a rhombus
(iv) a kite shaped figure triangle.
(v) a rectangle
(vi) a square
(vii) an isosceles

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 18.

Question 4.
Draw a triangle with :
(i) no line of symmetry,
(ii) only one line of symmetry,
(iii) exactly two lines of symmetry,
(iv) exactly three lines of symmetry,
(v) more than three lines of symmetry.
In each case, if possible, represent the line/ lines of symmetry by dotted lines. Also, write the special name of the triangle drawn.

Solution:
(i) Scalene triangle : It has no line of symmetry.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 19
(ii) Isosceles Triangle : It has one line of symmetry as shown.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 20
(iii) It is not possible.
(iv) Equilateral Triangle : It has three lines of symmetry as shown
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 21
(v) It is not possible.

Question 5.
Draw a quadrilateral with :
(i) no line of symmetry.
(ii) only one line of symmetry.
(iii) exactly two lines of symmetry.
(iv) exactly three lines of symmetry.
(v) exactly four lines of symmetry.
(vi) more than four lines of symmetry.
In each case, if possible, represent the line/ lines of symmetry by dotted lines. Also, write the special name of the quadrilateral drawn.

Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 22

Question 6.
Construct an equilateral triangle with each side 6 cm. In the triangle drawn, draw all the possible lines of symmetry.

Solution:
Steps of Construction :
(i) Draw a line segment BC = 6 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 23
(ii) With centres B and C and radius 6 cm, draw two arcs intersecting each other at A.
(iii) Join AB and AC
∆ ABC is the required equilateral triangle,
(iv) Draw the angle bisectors of ∠A, ∠B and ∠C.
These are the lines of symmetry which are three in numbers as the triangle is equilateral.

Question 7.
Construct a triangle ABC in which AB = AC = 5cin and BC = 5.6 cm. If possible, draw its lines of symmetry.
Solution:
Steps of Construction :
(i) Draw a line segment BC = 5.6 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 24
(ii) With centres B and C and radius 5 cm, draw two arcs intersecting each other at A.
(iii) Join AB and AC.
∆ ABC is an isosceles triangle.
(iv) Draw the bisector of ∠A. This is the only one line of symmetry as the triangle is an isosceles.

Question 8.
Construct a triangle PQR such that PQ = QR = 5 .5 cm and angle PQR = 90°. If possible, draw its lines of symmetry.
Solution:
∴ ∠PQR = 90°, and ∠P = ∠R
(opposite sides are equal)
∴∠P + ∠R = 90°
Hence ZP = ZR = \(\frac { 99\circ }{ 2 }\)  = 45°
Steps of Construction:
(i) Draw a line segment QR = 5.5 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 25
(ii) At Q, draw a ray making an angle of 90° and cut off QP = 5.5 cm.
(iii) Join PR.
∆PQR is an isosceles triangle.
(iv) Draw the angle bisector of ∠PQR. It is the line of symmetry. Since the triangle is an isosceles.
∴It has only one line of symmetry.

Question 9.
If possible, draw a rough sketch of a quadrilateral which has exactly two lines of symmetry.

Solution:
Since the quadrilateral has exactly two lilies of symmetry
∴It must be a rectangle or a rhombus
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 26

Question 10.
A quadrilateral ABCD is symmetric about its diagonal AC. Name tire sides of this quadrilateral which are equal.

Solution:
The quadrilateral ABCD is symmetric about its diagonal AC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 27
∴It must be a kite shaped.
Hence side AB = AD and BC = DC.

Symmetry Exercise 17B – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
In each figure, given below, find the image of the point P in the line AB :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 28
Solution:
Steps of Construction : Fig. (i) and (ii)
(i) From P, draw a perpendicular to the given line AB meeting it at O.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 29
(ii) Produce PO to P’ such that OP’ = PO.
P’ is the required image of P in AB.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 30

Question 2.
In each figure, given below, find the image of the line segment AB in the line PQ :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 31
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 32
Steps of Construction :
(i) From A and B, draw perpendiculars on PQ intersecting PQ at L and M.
(ii) Produce AL to A’ such that AL = LA’ and produce BM to B’ such that BM = MB’ A’B’ is the image of the line segment AB in PQ.

Question 3.
Complete the following table :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 33
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 34

Question 4.
A point P (7,3)|is reflected in x-axis to point P’. The point P’ is further reflected in v-axis to point P” Find :
(i) the co-ordinates of P’
(ii) the co-ordinates of P”
(iii) the image of P (7, 3) in origin.
Solution:
(i) Image of point P (7,3) when reflected in x-axis is P’ whose co-ordinates will be (7,-3)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 35
(ii) Image of point P’ (7,-3) when reflected in y-axis, is P” whose co-ordinates will be (- 7,-3)
(iii) The image of P (7, 3) in origin is P” whose co-ordinates are (- 7, – 3).

Question 5.
A point A (- 5, 4) is reflected in y-axis to point B. The point B is further reflected in origin to point C. find :
(i) the co-ordinates of B
(ii) the co-ordinates of C
(iii) the image of A (- 5, 4) in x-axis.
Solution:
(i) Image of point A (- 5,4) when reflected in y-axis is B whose co-ordinates will be (5,4)
(ii) Image of B (5, 4) when reflected in origin is C whose co-ordinates will be (- 5, – 4)
(iii) Image of A (- 5,4) in x-axis is C whose co-ordinates are (- 5, – 4)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 36

Question 6.
The point P (3, – 8) is reflected in origin to point Q. The Point Q is further reflected in x-axis to point R. Find :
(i) the co-ordinates of Q
(ii) the co-ordinates of R
(iii) the image of P (3, – 8) in y-axis.
Solution:
(i) The image of the given point P (3, – 8) when reflected in origin is Q whose co-ordinates will be (- 3, 8).
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 37
(ii) The image of Q (- 3, 8) when reflected in x-axis is R whose co-ordinates will be (-3,-8)
(iii) Tlie image of P (3, 8) in y-axis is R whose co-ordinates are (- 3, – 8).

Question 7.
Each of the points A (3, 0), B (7, 0), C (- 8, 0), D (- 7, 0) and E (0, 0) is reflected in x-axis to points A’, B’, C’, D’ and E’ respectively. Write the co-ordinates of each of the image points A’, B’, C’, D’ and E’.
Solution:
The points are given :
A (3, 0), B (1, 0), C (-8, 0), D (- 7, 0) and E (0, 0)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 38
This images will be when reflected in x-axis. A’ (3, 0), B’ (7, 0), C’ (- 8, 0) D’ (- 7, 0) and E’ (0, 0) as the given points lie on x-axis.

Question 8.
Each of the points A (0, 4), B (0, 10), C (0, – 4), D (0, – 6) and E (0, 0) is reflected in y-axis to points A’, B’, C’, D’ and E’ respectively. Write the co-ordinates of each of the image points A’, B’, C’, D’ and E’.
Solution:
The given points
A (0, 4), B (0, 10), C (0, – 4), D (0, – 6) and E (0, 0) are reflected in y-axis. The co – ordinates of their images will be A’ (0, 4), B’ (0, 10), C’ (0, – 4) D’ (0, – 6) and E’ (0, 0) as they all lie on y-axis.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 39

Question 9.
Each of the points A (0, 7), B (8. 0), C (0, -5), D (- 7, 0) and E (0, 0) are reflected in origin to points A’, B’, C’, D’ and E’ respectively. Write the co-ordinates of each of the image points A’, B’, C’, D’ and E’.
Solution:
The points A (0, 7), B (8, 0), C (0, – 5). D (- 7.0) and E (0,0) are reflected in origin.
So, the co-ordinates of their images will be A’ (0,-7), B’ (- 8, 0), C’ (0,5), D’ (7, 0) and E’ (0, 0)
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 40

Question 10.
Mark points A (4, 5) and B (- 5, 4) on a graph paper. Find A’, the image of A in x-axis and B’, the image of B in x-axis.
Mark A’ and B’ also on the same graph paper.
(ii) Join AB and A’ B’ and
find if AB = A’ B’ ?
Solution:
The given points :
A (4, 5) and B (- 5, 4) have been marked on the graph.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 41
The image of A in x-axis A (4, – 5) and image of B in x-axis is B’ (-5, -4) which have been also plotted on the same graph.
AB and A’ B’ are joined. We see that AB = A’ B’.

Question 11.
Mark points A (6, 4) and B (4, – 6) on a graph paper.
Find A’, the image of A in y-axis and B’, the image of B in y-axis. Mark A’ and B’ also on the same graph paper.
Solution:
The given points are
A (6, 4) and B (4, – 6)
The images of A and B is y-axis are A’ (- 6, 4) and B’ (- 4, – 6) respectively as shown in the same graph.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 42

Question 12.
Mark points A (- 6, 5) and B (- 4, – 6) on a graph paper. Find A’, the image of A in origin and B’, the image of B in origin. Mark A’ and B’ also on the same graph paper. Join AB and A’ B’. Is AB = A’ B’ ?
Solution:
The given points are A (- 6, 5) and B (- 4, – 6). The images of A and B in the origin are A’ and B’ where co-ordinates are A’ (6, – 5) and B’ (4, 6) which have been plotted on the same graph.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 43
AB and A’ B’ are joined we see that AB = A’ B’.

Symmetry Exercise 17C – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
How many lines of symmetry does a rhombus have?
Solution:
It has two lines of symmetry.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 44

Question 2.
What is the order of rotational symmetry of a rhombus?
Solution:
The order of rotational symmetry can be defined as the number of times that a shape appears exactly the same during a full 360° rotation. Order of rotational symmetry of a rhombus is 2.

Question 3.
Show that each of the following figures has two lines of symmetry and a rotational.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 45
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 46

Question 4.
Name a figure that has a line of symmetry but does not have any roational symmetry.
Solution:
Isosceles triangle has only line symmetry and no rotational symmetry.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 47

Question 5.
In each of the following figures, draw all possible lines of symmetry and also write the order of rotational symmetry:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 48
Solution:
(i) It has no line of symmetry and order of rotational symmetry is 3
(ii) It has 2 line of symmetry and order of rotational symmetry is 4.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 49
(iii) It has 2 line of symmetry and order of rotational symmetry is 2.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 50
(iv) It has no line of symmetry and order of rotational symmetry is 0.
(v) It has 1 line of symmetry and order of rotational symmetry is 0.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 17 Symmetry imagev - 51

 

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids (Representing 3-D in 2-D)

Selina Publishers Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids (Representing 3-D in 2-D)

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Recognition of solids Exercise 18 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Identify the nets which can be used to form cubes
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 1
Solution:
Nets for a cube are (ii) , (iii) and (v).

Question 2.
Draw at least three different nets for making cube.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 2

Question 3.
The figure, given below, shows shadows of some 3D objects, when seen under the lamp of an overhead projector :
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 3
In each case, name the object.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 4

Question 4.
Using Euler’s formula, find the values of a, b, c and d.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 5
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 6
(i) a + 6 – 12 = 2 ⇒ a = 2 – 6 + 12 = 14 – 6 = 8
(ii) b + 5- 9 = 2 ⇒6 = 2 + 9- 5 = 6
(iii) 20+ 12 — c = 2 ⇒32 – c = 2 ⇒ c = 32-2 ⇒ c = 30
(iv) 6 + d-12=2 ⇒ d – 6 = 2 ⇒ d = 2 + 6 = 8

Question 5.
Dice are cubes with dot or dots on each face. Opposite faces of a die always have a total of seven on them.
Below are given two nets to make dice (cube), the numbers inserted in each square indicate the number of dots in it.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 7
Insert suitable numbers in each blank so that numbers in opposite faces of the die have a total of seven dots.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 8

Question 6.
The following figures represent nets of some solids. Name the solids
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 9
Solution:
The given nets are of the solid as given below :
(i) Cube
(ii) Cuboid

Question 7.
Draw a map of your class room using proper scale and symbols for different objects.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 10

Question 8.
Draw a map of your school compound using proper scale and symbols for various features like play ground, main building, garden, etc.
Solution:
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 18 Recognition of Solids image - 11

Question 9.
In the map of India, the distance between two cities is 13.8 cm.
Taking scale : 1 cm = 12 km, find the actual distance between these two cities.
Solution:
The scale for a map is given to be 1 cm = 12 km
The distance between these two cities = 13.8 cm on the map
∴ Actual distance between these two cities
= 12 x 13.8 km = 165.6 km

Selina Concise Mathematics class 7 ICSE Solutions – Profit, Loss and Discount

Selina Concise Mathematics class 7 ICSE Solutions – Profit, Loss and Discount

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APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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POINTS TO REMEMBER

  1. The Cost Price (C.P.) of an article is the price at which the article is bought.
  2. The Selling Price (S.P.) of an article is the price at which the article is sold.
  3. If Selling Price of an article is more than its cost price ; it is sold at a profit (gain)
    Profit = Selling Price-Cost Price
    i.e., Profit (gain) = S.P. – C.P. and S.P. = C.P. + Gain
  4. If Selling Price of an article is less than its cost price ; it is sold at a loss.
    Loss = Cost Price – Selling Price
    i.e., Loss = C.P. – S.P. and S.P. = C.P. — Loss
  5. Profit percent and loss percent are always calculated on cost price (C.P.) only.
    i.e., (i) Profit % = \(\frac { Profit }{ C.P. }\) x 100% and
    (ii)
    Loss % = \(\frac {Loss }{ C.P. }\) x 100%
  6. Selling Price = Marked price – Discount
    i.e., S.P. = M.P.—(piscount
    Note : (i) Discount is calculated on marked price (M.P.)
    (ii) Marked price is also written as List price.

EXERCISE 9 (A)

Question 1.
Find the gain or loss percent, if
(i) C.P. = Rs. 200 and S.P.: = Rs. 224
(ii) C.P. = Rs. 450 and S.P. = Rs. 400
(iii) C.P. = Rs. 550 and gain = Rs . 22
(iv) CP. = Rs. 216 and loss = Rs. 72
(v) S.P. = Rs. 500 and loss : = Rs. 100
(vi) S.P. = Rs. 12 and profit = Rs. 4
(vii) C.P. = Rs. 5 and gain = 60 P

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 1
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 2
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 3

Question 2.
Find the selling price, if:
(i) C.P. = Rs. 500 and gain = 25%
(ii) C.P. = Rs. 60 and loss = 12 1/2%
(iii) C.P. = Rs. 150 and loss = 20%
(iv) C.P. = Rs. 80 and gain = 2.5%

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 4
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 5

Question 3.
Rohit bought a tape-recorder for Rs. 1,500 and sold it for Rs. 1,800. Calculate his profit or loss percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 6

Question 4.
An article bought for Rs. 350 is sold at a profit of 20%. Find its selling price.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 7

Question 5.
An old machine is bought for Rs. 1,400 and is sold at a loss of 15%. Find its selling price.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 8

Question 6.
Oranges are bought at 5 for Rs. 10 and sold at 6 for Rs. 15. Find profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 53

Question 7.
A certain number of articles are bought at 3 for Rs. 150 and all of them are sold at 4 for Rs. 180. Find the loss or gain as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 10

Question 8.
A vendor bought 120 sweets at 20 p each. In his house, 18 were consumed and he sold the remaining at 30 p each. Find his profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 11

Question 9.
The cost price of an article is Rs. 1,200 and selling price is \(\frac { 5 }{ 4 }\) times of its cost price. Find:
(i) selling price of the article
(ii) profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 12

Question 10.
The selling price of an article is Rs. 1,200 and cost price is \(\frac { 5 }{ 4 }\) times of its selling price,
find :
(i) cost price of the article ;
(ii) profit or loss as percent.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 13

EXERCISE 9 (B)

Question 1.
Find the cost price, if:
(i) S.P. = Rs. 21 and gain = 5%
(ii) S.P. = Rs. 22 and loss = 12%
(iii) S.P. = Rs. 340 and gain = Rs. 20
(iv) S.P. = Rs. 200 and loss = Rs. 50
(v) S.P. = Re. 1 and loss = 5 p.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 14

Question 2.
By selling an article for Rs. 810, a loss of 10 percent is suffered. Find its cost price.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 15

Question 3.
By selling a scooter for Rs. 9,200, a man gains 15%. Find the cost price of the scooter.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 16

Question 4.
On selling an article for Rs. 2,640, a profit of 10 percent is made. Find
(i) cost price of the article
(ii) new selling price of it, in order to gain 15%

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 17

Question 5.
A T.V. set is sold for Rs. 6800 at a loss of 15%. Find
(i)cost price of the T.V. set.
(ii)new selling price of it, in order to gain 12%

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 18

Question 6.
A fruit seller bought mangoes at Rs. 90 per dozen and sold them at a loss of 8 percent. How much will a customer pay for.
(i) one mango
(ii) 40 mangoes

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 19

Question 7.
By selling two transistors for Rs. 00 each, a shopkeeper gains 20 percent on one transistor and loses 20 percent on the other.
Find :
(i) C.P. of each transistor

(ii) total C.P. and total S.P. of both the transistors
(iii) profit or loss percent on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 20
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 21

Question 8.
Mangoes are bought at 20 for Rs. 60. If
they are sold at 33\(\frac { 1 }{ 3 }\) percent profit.
Find:

(i) selling price of each mango.
(ii) S.P. of 8 mangoes.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 22

Question 9.
Find the cost price of an article, which is sold for Rs. 4050 at a loss of 10%. Also, find the new selling price of the article which must give a profit of 8%.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 23
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 24

Question 10.
By selling an article for ₹825, a man loses \(\frac { 1 }{ 3 }\) equal to j of its selling price.
Find :

(i) the cost price of the article,
(ii) the profit percent or the loss percent made, if the same article is sold for ₹1265.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 25

Question 11.
Find the loss or gain as percent, if the C.P. of 10 articles, all of the same kind, is equal to S.P. of 8 articles.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 26
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 27

Question 12.
Find the loss or gain as percent, if the C.P. of 8 articles, all of the same kind, is equal to S.P. of 10 articles.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 28

Question 13.
The cost price of an article is 96% of its selling price. Find the loss or the gain as percent on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 29
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 30

Question 14.
The selling price of an article is 96% of its cost price. Find the loss or the gain as percent on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 31

Question 15.
Hundred oranges are bought for ₹350 and all of them are sold at the rate of ₹48 per dozen. Find the profit percent or loss percent made.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 32

Question 16.
Oranges are bought at 100 for ?80 and all of them are sold at ₹80 for ₹100. Find the loss or gain as percent in this transaction.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 33

Question 17.
An article is bought for ₹5,700 and ₹1,300 is spent on its repairing, transportion, etc. For how much should this article be sold in order to gain 20% on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 34

EXERCISE 9 (C)

Question 1.
A machine is marked at ₹5000 and is sold at a discount of 10%. Find the selling price of the machine.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 35

Question 2.
shopkeeper marked a dinner set for ₹1000. He sold it at ₹900, what percent discount did he give ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 36

Question 3.
A pair of shoes marked at ₹320, are sold at a discount of 15 percent.
Find :
(i) discount
(ii) selling price of the shoes.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 37

Question 4.
The list price of an article is ₹450 and it is sold for ₹360.
Find :
(i) discount
(ii) discount percent

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 38

Question 5.
A shopkeeper buys an article for₹300. He increases its price by 20% and then gives 10% discount on the new price. Find:
(i) the new price (marked price) of the article.
(ii) the discount given by the shopkeeper.
(iii) the selling price.
(iv) profit percent made by the shopkeeper.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 39
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 40

Question 6.
A car is marked at Rs. 50,000. The dealer gives 5% discount on first Rs. 20,000 and 2% discount on the remaining Rs. 30,000.
Find :
(i) the total discount.
(ii) the price charged by the dealer.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 41

Question 7.
A dealer buys a T.V. set for Rs. 2500. He marks it at Rs. 3,200 and then gives a discount of 10% on it.
Find :
(i) the selling price of the T.V. set
(ii) the profit percent made by the dealer.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 42

Question 8.
A sells his goods at 15% discount. Find the price of an article which is sold for Rs. 680.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 43
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 44

Question 9.
A shopkeeper allows 20% discount on the marked price of his articles. Find the marked price of an article for which he charges Rs. 560.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 45

Question 10.
An article is bought for Rs. 1,200 and Rs. 100 is spent on its transportation, etc.
Find :
(i) the total C.P. of the article.
(ii) the selling price of it in order to gain 20% on the whole.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 46

Question 11.
40 pens are bought at 4 for Rs. 50 and all of them are sold at 5 for Rs. 80
Find :
(i) C.P. of one pen.
(ii) S/P. of one pen.
(iii) Profit made by selling one pen.
(iv) Profit percent made by selling one pen.
(v) C.P. of 40 pens
(vi) S.P. of 40 pens.
(vii) Profit made by selling 40 pens.
(viii) Profit percent made by selling 40 pens. Are the results of parts (iv) and (viii) same? What conclusion do you draw from the above result ?

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 47
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 48

Question 12.
The C.P. of 5 identical articles is equal to S.P. of 4 articles. Calculate the profit percent or loss percent made if all the articles bought are sold.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 49

Question 13.
The C.P. of 8 pens is same as S.P. of 10 pens. Calculate the profit or loss percent made, if all the pens bought are considered to be sold

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 50

Question 14.
A certain number of articles are bought at Rs. 450 per dozen and all of them are sold at a profit of 20%. Find the S.P. of:
(i) one article
(ii)seven articles.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 51

Question 15.
An article is marked 60% above the cost price and sold at 20% discount. Find the profit percent made.

Solution:
Selina Concise Mathematics class 7 ICSE Solutions - Profit, Loss and Discount image - 52

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

Selina Publishers Concise Maths Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem

ICSE SolutionsSelina ICSE SolutionsML Aggarwal Solutions
APlusTopper.com provides step by step solutions for Selina Concise ICSE Solutions for Class 7 Mathematics. You can download the Selina Concise Mathematics ICSE Solutions for Class 7 with Free PDF download option. Selina Publishers Concise Mathematics for Class 7 ICSE Solutions all questions are solved and explained by expert mathematic teachers as per ICSE board guidelines.

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Pythagoras Theorem Exercise 16 – Selina Concise Mathematics Class 7 ICSE Solutions

Question 1.
Triangle ABC is right-angled at vertex A. Calculate the length of BC, if AB = 18 cm and AC = 24 cm.

Solution:
Given : ∆ABC right angled at A and AB = 18 cm, AC = 24 cm.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 1

To find : Length of BC.
According to Pythagoras Theorem,
BC2 = AB2 + AC2
= 182 + 242 = 324 + 576 = 900
∴BC = \(\sqrt { 900 }\) = \(\sqrt { 30 x 30 }\)= 30 cm

Question 2.
Triangle XYZ is right-angled at vertex Z. Calculate the length of YZ, if XY = 13 cm and XZ = 12 cm.

Solution:
Given : ∆XYZ right angled at Z and XY = 13 cm, XZ = 12 cm.
To find : Length of YZ.
According to Pythagoras Theorem,
XY2 = XZ2 + YZ2
132 = 122 + YZ2
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 2

169= 144 +YZ2
169- 144 = YZ2
25 = YZ2
∴YZ = \(\sqrt { 25 }\)cm \(\sqrt { 5×5 }\) = 5 cm

Question 3.
Triangle PQR is right-angled at vertex R. Calculate the length of PR, if:
PQ = 34 cm and QR = 33.6 cm.

Solution:
Given : ∆PQR right angled at R and PQ = 34 cm, QR = 33.6 cm.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 3

To find : Length of PR.
According to Pythagoras Theorem,
PR2 + QR2 = PQ2
PR2 + 33.62 = 342
PR2+ 1128.96= 1156
PR2 = 1156- 1128.96
∴ PR = \(\sqrt { 27.04 }\) = 5.2 cm

Question 4.
The sides of a certain triangle are given below. Find, which of them is right-triangle
(i) 16 cm, 20 cm and 12 cm
(ii) 6 m, 9 m and 13 m

Solution:
(i) 16 cm, 20 cm and 12 cm
The given triangle will be a right-angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
i.e., If (20)2 = (16)2 = (12)2
(20)2 = (16)2 + (12)2
400 = 256 + 144
400 = 400
So, the given triangle is right angled.
(ii) 6 m, 9 m and 13 m
The given triangle will be a right-angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
i.e., If (13)2 = (9)2 + (6)2
169 = 81+36 169 ≠ 117
So, the given triangle is not right angled.

Question 5.
In the given figure, angle BAC = 90°, AC = 400 m and AB = 300 m. Find the length of BC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 4
Solution:
AC = 400 m
AB = 300 m
BC = ?
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 5
According to Pythagoras Theorem,
BC2 = AB2 + AC2
BC2 = (300)2 + (400)2
BC2 = 90000 + 160000
BC2 = 250000
BC = \(\sqrt { 250000 }\)= 500 m

Question 6.
In the given figure, angle ACP = ∠BDP = 90°, AC = 12 m, BD = 9 m and PA= PB = 15 m. Find:

(i) CP
(ii) PD
(iii) CD
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 6

Solution:
Given : AC = 12 m
BD = 9 m
PA = PB= 15 m
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 7
(i) In right angle triangle ACP
(AP)2 = (AC)2 + (CP)2
152 = 122 + CP2
225 = 144 + CP2
225 – 144 = CP2
81 =CP
\(\sqrt { 81 }\) =CP
∴ CP = 9 m
(ii) In right angle triangle BPD
(PB)2 = (BD)2 + (PD)2
(15)2 = (9)2 + PD2
225 = 81 + PD2
225-81 = PD2
144 = PD2
\(\sqrt { 144 }\)=PD                                               ‘
∴ PD = 12 m
(iii) CP = 9 m
PD = 12 m
∴ CD = CP + PD
= 9+ 12 = 21 m

Question 7.
In triangle PQR, angle Q = 90°, find :
(i) PR, if PQ = 8 cm and QR = 6 cm
(ii) PQ, if PR = 34 cm and QR = 30 cm

Solution:
(i) Given:
PQ = 8 cm
QR = 6 cm
PR = ?
∠PQR = 90°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 8
According to Pythagoras Theorem,
(PR)2 = (PQ)2 + (QR)2
PR2 = 82 + 62
PR2 = 64 + 36
PR2 = 100
∴ PR = \(\sqrt { 100 }\)= 10 cm
(ii) Given :
PR = 34 cm
QR = 30 cm
PQ = ?
∠PQR = 90°
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 9
According to Pythagoras Theorem,
(PR)2 = (PQ)2 + (QR)2
(34)2 = PQ2 + (30)2
1156 = PQ2 + 900
1156-900 = PQ2
256 = PQ2
∴ PQ = 16 cm

Question 8.
Show that the triangle ABC is a right-angled triangle; if:
AB = 9 cm, BC = 40 cm and AC = 41 cm

Solution:
AB = 9 cm
CB = 40 cm
AC = 41 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 10
The given triangle will be a right angled triangle if square of its largest side is equal to the sum of the squares on the other two sides.
According to Pythagoras Theorem,
(AC)2 = (BC)2 + (AB)2
(41)2 = (40)2 + (9)2
1681 = 1600 + 81
1681 = 1681
Hence, it is a right-angled triangle ABC.

Question 9.
In the given figure, angle ACB = 90° = angle ACD. If AB = 10 m, BC = 6 cm and AD = 17 cm, find :
(i) AC
(ii) CD
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 11

Solution:
Given:
∆ABD
∠ACB = ∠ACD = 90°
and AB = 10 cm, BC = 6 cm and AD = 17 cm
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 12
To find:
(i) Length of AC
(ii) Length of CD
Proof:
(i) In right-angled triangle ABC
BC = 6 cm, AB = 110 cm
According to Pythagoras Theorem,
AB2 = AC2 + BC2
(10)2 = (AC)2 + (6)2
100 = (AC)2 + 36
AC2 = 100-36 = 64 cm
AC2 = 64 cm
∴ AC = \(\sqrt { 8×8 }\) = 8 cm
(ii) In right-angle triangle ACD
AD = 17 cm, AC = 8 cm
According to Pythagoras Theorem,
(AD)2 = (AC)2 + (CD)2
(17)2 = (8)2 + (CD)2
289 – 64 = CD2
225 = CD2
CD =\(\sqrt { 15×15 }\) = 15 cm

Question 10.
In the given figure, angle ADB = 90°, AC = AB = 26 cm and BD = DC. If the length of AD = 24 cm; find the length of BC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 13

Solution:
Given:
∆ABC
∠ADB = 90° and AC = AB = 26 cm
AD = 24 cm
To find : Length of BC In right angled ∆ADC
AB = 26 cm, AD = 24 cm
According to Pythagoras Theorem,
(AC)2 = (AD)2 + (DC)2
(26)2 = (24)2 + (DC)2
676 = 576 + (DC)2
⇒ (DC)2 = 100
⇒ DC = \(\sqrt { 100 }\) = 10 cm
∴ Length of BC = BD + DC
= 10 + 10 = 20 cm

Question 11.
In the given figure, AD = 13 cm, BC = 12 cm, AB = 3 cm and angle ACD = angle ABC = 90°. Find the length of DC.

Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 14

Solution:
Given :
∆ACD = ∠ABC = 90°
and AD = 13 cm, BC = 12 cm, AB = 3 cm
To find : Length of DC.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 15

(i)In right angled ∆ABC
AB = 3 cm, BC = 12 cm
According to Pythagoras Theorem,
(AC)2 = (AB)2 + (BC)2
(AC)2 = (3)2 + (12)2
(AC) = \(\sqrt { 9+144 }\) = \(\sqrt { 153 }\) cm

 (ii) In right angled triangle ACD
AD = 13 cm, AC =\(\sqrt { 153 }\)
According to Pythagoras Theorem,
DC2 = AB2-AC2
DC2= 169-153
DC = \(\sqrt { 16 }\) = 4 cm
∴ Length of DC is 4 cm

Question 12.
A ladder, 6.5 m long, rests against a vertical wall. Ifthe foot of the ladcler is 2.5 m from the foot of the wall, find upto how much height does the ladder reach?

Solution:
Given :
Length of ladder = 6.5 m
Length of foot of the wall = 2.5 m
To find : Height AC According to Pythagoras Theorem,
(BC)2 = (AB)2 + (AC)2
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 16

(6.5)2 = (2.5)2 + (AC)2
42.25 = 6.25 + AC2
AC2 = 42.25 – 6.25 = 36 m
AC = \(\sqrt { 6×6 }\) = 6 m
∴ Height of wall = 6 m

Question 13.
A boy first goes 5 m due north and then 12 m due east. Find the distance between the initial and the final position of the boy.

Solution:
Given : Direction of north = 5 m i.e. AC Direction of east = 12 m i.e. AB
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 17

To find: BC
According to Pythagoras Theorem,
In right angled AABC
(BC)2 = (AC)2 + (AB)2
(BC)2 = (5)2 + (12)2
(BC)2 = 25 + 144
(BC)2 = 25 + 144
(BC)2= 169
∴ BC = \(\sqrt { 169 }\) = \(\sqrt { 13×13 }\) = 13 m

Question 14.
Use the information given in the figure to find the length AD.
Selina Concise Mathematics Class 7 ICSE Solutions Chapter 16 Pythagoras Theorem image - 18

Solution:
Given :
AB = 20 cm
∴AO =\(\frac { AB }{ 2 }\) = \(\frac { 20 }{ 2 }\) =10cm
BC = OD = 24 cm
To find : Length of AD
In right angled triangle
AOD (AD)2 = (AO)2 + (OD)2
(AD)2 = (10)2 + (24)2
(AD)2 = 100 + 576
(AD)2 = 676
∴ AD =  \(\sqrt { 26×26 }\)
AD = 26 cm