CBSE

RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions

Exercise 11D

Very-Short and Short-Answer Questions
Question 1.
Solution:
(3y – 1), (3y + 5) and (5y+ 1) are in AP
(3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)
⇒ 2 (3y + 5) = (5y + 1) + (3y – 1)
⇒ 6y + 10 = 8y
⇒ 8y – 6y = 10
⇒ 2y = 10
⇒ y = 5
y = 5

Question 2.
Solution:
k, (2k – 1) and (2k + 1) are the three successive terms of an AP.
(2k – 1) – k = (2k + 1) – (2k – 1)
⇒ 2 (2k – 1) = 2k + 1 + k
⇒ 4k – 2 = 3k + 1
⇒ 4k – 3k = 1 + 2
⇒ k = 3
k = 3

Question 3.
Solution:
18, a, (b – 3) are in AP
⇒ a – 18 = b – 3 – a
⇒ a + a – b = -3 + 18
⇒ 2a – b = 15

Question 4.
Solution:
a, 9, b, 25 are in AP.
9 – a = b – 9 = 25 – b
b – 9 = 25 – b
⇒ b + b = 22 + 9 = 34
⇒ 2b = 34
⇒ b= 17
a – b = a – 9
⇒ 9 + 9 = a + b
⇒ a + b = 18
⇒ a + 17 = 18
⇒ a = 18 – 17 = 1
a = 18, b= 17

Question 5.
Solution:
(2n – 1), (3n + 2) and (6n – 1) are in AP
(3n + 2) – (2n – 1) = (6n – 1) – (3n + 2)
⇒ (3n + 2) + (3n + 2) = 6n – 1 + 2n – 1
6n + 4 = 8n – 2
⇒ 8n – 6n = 4 + 2
⇒ 2n = 6
⇒ n = 3
and numbers are
2 x 3 – 1 = 5
3 x 3 + 2 = 11
6 x 3 – 1 = 17
i.e. (5, 11, 17) are required numbers.

Question 6.
Solution:
Three digit numbers are 100 to 990 and numbers which are divisible by 7 will be
105, 112, 119, 126, …, 994
Here, a = 105, d= 7, l = 994
T_n = (l) = a + (n – 1) d
⇒ 994 = 105 + (n – 1) x 7
⇒ 994 – 105 = (n – 1) 7
⇒ (n – 1) x 7 = 889
⇒ n – 1 = 127
⇒ n = 127 + 1 = 128
Required numbers are 128

Question 7.
Solution:
Three digit numbers are 100 to 999
and numbers which are divisible by 9 will be
108, 117, 126, 135, …, 999
Here, a = 108, d= 9, l = 999
T_n (l) = a + (n – 1) d
⇒ 999 = 108 + (n – 1) x 9
⇒ (n – 1) x 9 = 999 – 108 = 891
⇒ n – 1 = 99
⇒ n = 99 + 1 = 100

Question 8.
Solution:
Sum of first m terms of an AP = 2m² + 3m
S_m = 2m² + 3m
S₁ = 2(1)² + 3 x 1 = 2 + 3 = 5
S₂ = 2(2)² + 3 x 2 = 8 + 6=14
S₃ = 2(3)² + 3 x 3 = 18 + 9 = 27
Now, T₂ = S₂ – S₁ = 14 – 5 = 9
Second term = 9

Question 9.
Solution:
AP is a, 3a, 5a, …
Here, a = a, d = 2a

Question 10.
Solution:
AP 2, 7, 12, 17, …… 47
Here, a = 2, d = 7 – 2 = 5, l = 47
nth term from the end = l – (n – 1 )d
5th term from the end = 47 – (5 – 1) x 5 = 47 – 4 x 5 = 47 – 20 = 27

Question 11.
Solution:
AP is 2, 7, 12, 17, …
Here, a = 2, d = 7 – 2 = 5
a_n = a + (n – 1) d = 2 + (n – 1) x 5 = 2 + 5n – 5 = 5n – 3
Now, a₃₀ = 2 + (30 – 1) x 5 = 2 + 29 x 5 = 2 + 145 = 147
and a₂₀ = 2 + (20 – 1) x 5 = 2 + 19 x 5 = 2 + 95 = 97
a₃₀ – a₂₀ = 147 – 97 = 50

Question 12.
Solution:
T_n = 3n + 5
T_n-1 = 3 (n – 1) + 5 = 3n – 3 + 5 = 3n + 2
d = T_n – T_n-1 = (3n + 5) – (3n + 2) = 3n + 5 – 3n – 2 = 3
Common difference = 3

Question 13.
Solution:
T_n = 7 – 4n
T_n-1 = 7 – 4(n – 1) = 7 – 4n + 4 = 11 – 4n
d = T_n – T_n-1 = (7 – 4n) – (11 – 4n) = 7 – 4n – 11 + 4n = -4
d = -4

Question 14.
Solution:
AP is √8, √18, √32, …..
⇒ √(4 x 2) , √(9 x 2) , √(16 x 2), ………

Question 15.
Solution:

Question 16.
Solution:
AP is 21, 18, 15, …n
Here, a = 21, d = 18 – 21 = -3, l = 0
T_n (l) = a + (n – 1) d
0 = 21 + (n – 1) x (-3)
0 = 21 – 3n + 3
⇒ 24 – 3n = 0
⇒ 3n = 24
⇒ n = 8 .
0 is the 8th term.

Question 17.
Solution:
First n natural numbers are 1, 2, 3, 4, 5, …, n
Here, a = 1, d = 1

Question 18.
Solution:
First n even natural numbers are 2, 4, 6, 8, 10, … n
Here, a = 2, d = 4 – 2 = 2

Question 19.
Solution:
In an AP
First term (a) = p
and common difference (d) = q
T₁₀ = a + (n – 1) d = p + (10 – 1) x q = (p + 9q)

Question 20.
Solution:

Question 21.
Solution:
2p + 1, 13, 5p – 3 are in AP, then
13 – (2p + 1) = (5p – 3) – 13
⇒ 13 – 2p – 1 = 5p – 3 – 13
⇒ 12 – 2p = 5p – 16
⇒ 5p + 2p = 12 + 16
⇒ 7p = 28
⇒ p = 4
P = 4

Question 22.
Solution:
(2p – 1), 7, 3p are in AP, then
⇒ 7 – (2p – 1) = 3p – 7
⇒ 7 – 2p + 1 = 3p – 7
⇒ 7 + 1 + 7 = 3p + 2p
⇒ 5p = 15
⇒ p = 3
P = 3

Question 23.
Solution:

Question 24.
Solution:

d = T₂ – T₁ = 14 – 8 = 6
Common difference = 6

Question 25.
Solution:

Question 26.
Solution:

Question 27.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions are helpful to complete your math homework.

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RS Aggarwal Solutions Class 11 Chapter 11 Arithmetic Progressions MCQS

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions

Choose the correct answer in each of the following questions.
Question 1.
Solution:

Question 2.
Solution:

Question 3.
Solution:

Question 4.
Solution:

Question 5.
Solution:

Question 6.
Solution:

Question 7.
Solution:

Question 8.
Solution:

Question 9.
Solution:

Question 10.
Solution:

Question 11.
Solution:

Question 12.
Solution:

Question 13.
Solution:

Question 14.
Solution:
Sum of first n odd natural numbers = (n)²
Sum of first 20 odd natural numbers = (20)² = 400 (c)

Question 15.
Solution:
First 40 positive integers divisible by 6 are 6, 12, 18, 24, … to 40 terms
Here, a = 6, d = 6, n = 40

Question 16.
Solution:

Question 17.
Solution:

Question 18.
Solution:
In an AP,
a₁₈ – a₁₄ = 32
Let a be the first term and d be the common difference, then

Question 19.
Solution:

Question 20.
Solution:

Question 21.
Solution:

Question 22.
Solution:

Question 23.
Solution:

Question 24.
Solution:

Question 25.
Solution:
Sum of first 16 terms of AP 10, 6, 2, …
Here, a = 10, d = 6 – 10 = -4, n = 16

Question 26.
Solution:

Question 27.
Solution:

Question 28.
Solution:
In an AP, T₁₇ = T₁₀ + 21
Let a be the first term and d be the common difference, then

Question 29.
Solution:

Question 30.
Solution:

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions are helpful to complete your math homework.

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RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions Ex 11B

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions

Exercise 11B

Question 1.
Solution:
(3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.
(4k – 6) – (3k – 2) = (k + 2) – (4k – 6)
⇒ 2(4k – 6) = (k + 2) + (3k – 2)
⇒ 8k – 12 = 4k + 0
⇒ 8k – 4k = 0 + 12
⇒ 4k = 12
k = 3

Question 2.
Solution:
(5x + 2), (4x – 1) and (x + 2) are in AP.
(4x – 1) – (5x + 2) = (x + 2) – (4x – 1)
⇒ 2(4x – 1) = (x + 2) + (5x + 2)
⇒ 8x – 2 = 6x + 2 + 2
⇒ 8x – 2 = 6x + 4
⇒ 8x – 6x = 4 + 2
⇒ 2x = 6
x = 3

Question 3.
Solution:
(3y – 1), (3y + 5) and (5y + 1) are the three consecutive terms of an AP.
(3y + 5) – (3y – 1) – (5y + 1) – (3y + 5)
⇒ 2(3y + 5) = 5y + 1 + 3y – 1
⇒ 6y + 10 = 8y
⇒ 8y – 6y = 10
⇒ 2y = 10
⇒ y = 5
y = 5

Question 4.
Solution:
(x + 2), 2x, (2x + 3) are three consecutive terms of an AP.
2x – (x + 2) = (2x + 3) – 2x
⇒ 2x – x – 2 = 2x + 3 – 2x
⇒ x – 2 = 3
⇒ x = 2 + 3 = 5
x = 5

Question 5.
Solution:
(a – b)², (a² + b²) and (a + b)² will be in AP.
If (a² + b²) – (a – b)² = (a + b)² – (a² + b²)
If (a² + b²) – (a² + b² – 2ab) = a² + b² + 2ab – a² – b²
2ab = 2ab which is true.
Hence proved.

Question 6.
Solution:
Let the three numbers in AP be
a – d, a, a + d
a – d + a + a + d = 15
⇒ 3a = 15
⇒ a = 5
and (a – d) x a x (a + d) = 80
a(a² – d²) = 80
⇒ 5(5² – d²) = 80
⇒ 25 – d² = 16
⇒ d² = 25 – 16 = 9 = (±3)²
d = ±3
Now, a = 5, d = +3
Numbers are 5 – 3 = 2
5 and 5 + 3 = 8
= (2, 5, 8) or (8, 5, 2)

Question 7.
Solution:
Let the three numbers in AP be a – d, a and a + d

Question 8.
Solution:
Sum of three numbers = 24
Let the three numbers in AP be a – d, a, a + d

Question 9.
Solution:
Let three consecutive in AP be a – d, a, a + d
a – d + a + a + d = 21
⇒ 3a = 21

Question 10.
Solution:
Sum of angles of a quadrilateral = 360°
Let d= 10
The first number be a, then the four numbers will be
a, a + 10, a + 20, a + 30
a + a + 10 + a + 20 + a + 30 = 360
4a + 60 = 360
4a = 360 – 60 = 300
Angles will be 75°, 85°, 95°, 105°

Question 11.
Solution:
Let the four numbers in AP be a – 3d, a – d, a + d, a + 3d, then
a – 3d + a – d + a + d + a + 3d = 28

Question 12.
Solution:
Let the four parts of 32 be a – 3d, a – d, a + d, a + 3d

Question 13.
Solution:
Let the three terms be a – d, a, a + d
a – d + a + a + d = 48
⇒ 3a = 48
⇒ a = 16
and (a – d) x a = (a + d) + 12
⇒ a(a – d) = 4 (a + d) + 12
⇒ 16 (16 – d) = 4(16 + d) + 12
⇒ 256 – 16d = 64 + 4d + 12 = 4d + 76
⇒ 256 – 76 = 4d + 16d
⇒ 180 = 20d
⇒ d = 9
Numbers are (16 – 9, 16), (16 + 9) or (7, 16, 25)

Hope given RS Aggarwal Solutions Class 10 Chapter 11 Arithmetic Progressions are helpful to complete your math homework.

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RS Aggarwal Solutions Class 10 Chapter 4 Triangles Ex 4D

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles

Exercise 4D

Question 1.
Solution:
For the given triangle to be right-angled, the sum of the squares of the two smaller sides must be equal to the square of the largest side.
(i) 9 cm, 16 cm, 18 cm
Longest side = 18
Now (18)² = 324
and (9)² + (16)² = 81 + 256 = 337
324 ≠ 337
It is not a right triangle.
(ii) 1 cm, 24 cm, 25 cm
Here longest side = 25 cm
(25)² = 625
and (7)² x (24)² = 49 + 576 = 625
625 = 625
It is a right triangle
(iii) 1.4 cm, 4.8 cm, 5 cm
Here longest side = 5 cm
(5)² = 25
and (1.4)² + (4.8)² = 1.96 + 23.04 = 25.00 = 25
25 = 25
It is a right triangle
(iv) 1.6 cm, 3.8 cm, 4 cm
Here longest side = 4 cm
(4 )² = 16
and (1.6)² + (3.8)² = 2.56 + 14.44 = 17.00 = 17
16 ≠ 17
It is not a right triangle
(v) (a- 1) cm, 2√a cm, (a + 1) cm
Here longest side = (a + 1) cm
(a + 1)² = a² + 2a + 1
and (a – 1)² + (2 √a )² = a² – 2a + 1 + 4a = a² + 2a + 1
a² + 2a + 1 = a² + 2a + 1
It is a right triangle.

Question 2.
Solution:
A man goes 80 m from O to east side and reaches A, then he goes 150 m due north from A and reaches B.
Join OB.

In right ∆OAB,
OB² = OA²+² (Pythagoras Theorem) = (80)² + (150)² = 6400 + 22500 = 28900
⇒ OB = √28900 = 170

He is 170 m away from the starting point.

Question 3.
Solution:
A man goes 10 m due south from O and reaches A and then 24 m due west from A and reaches B.
Join OB.

Question 4.
Solution:
Length of a ladder = 13 m
Height of the window = 12 m
Distance between the foot of the ladder and building.
In the figures,

AB is ladder, A is window of building AC
AB² = AC² + BC² (Pythagoras Theorem)
⇒ (13)² = (12)² + x²
⇒ 169 = 144 + x²
⇒ x² = 169 – 144 = 25 = (5)²
x = 5
Hence, distance between foot of ladder and building = 5 m.

Question 5.
Solution:
Let length of ladder AB = x m
Height of window AC = 20 m
and distance between the foot of the ladder and the building (BC) = 15 m

AB² = AC² + BC² (Pythagoras Theorem)
⇒ x² = 20² + 15² = 400 + 225 = 625 = (25)²
x = 25
Length of ladder = 25 m

Question 6.
Solution:
Height of first pole AB = 9 m
and of second pole CD = 14 m

Let distance between their tops CA = x m
From A, draw AE || BD meeting CD at E.
Then EA = DB = 12 m CE = CD – ED = CD – AB = 14-9 = 5 m
In right ∆AEC,
AC² = AE² + CE² = 122 + 52 = 144 + 25 = 169 = (13)²
AC = 13
Distance between their tops = 13 m

Question 7.
Solution:
Height of the pole AB = 18 m
and length of wire AC = 24 m

Distance between the base of the pole and other end of the wire
BC = x m (suppose)
In right ∆ABC,
AC² = AB² + BC² (Pythagoras Theorem)
(24)² = (18)² + x²
⇒ 576 = 324 + x²
⇒ x² = 576 – 324 = 252

Question 8.

Solution:
In ∆PQR, O is a point in it such that
OP = 6 cm, OR = 8 cm and ∠POR = 90°
PQ = 24 cm, QR = 26 cm
To prove : ∆PQR is a right angled.
In ∆POR, ∠O = 90°
PR² = PO² + OR² = (6)² + (8)² = 36 + 64 = 100 = (10)²
PR = 10
Greatest side QR is 26 cm
QR² = (26)² = 676
and PQ² + PR² = (24)² + (10)² = 576 + 100 = 676
676 = 676
QR² = PQ² + PR²
∆PQR is a right angled triangle and right angle at P.

Question 9.
Solution:
In isosceles ∆ABC, AB = AC = 13 cm
AL is altitude from A to BC
and AL = 5 cm

Now, in right ∆ALB
AB² = AL² + BL²
(13)² = (5)² + BL²
⇒ 169 = 25 + BL²
⇒ BL² = 169 – 25 = 144 = (12)²
BL = 12 cm
L is mid point of BC
BC = 2 x BC = 2 x 12 = 24 cm

Question 10.
Solution:
In an isosceles ∆ABC in which
AB = AC = 2a units, BC = a units

Question 11.
Solution:
∆ABC is an equilateral triangle
and AB = BC = CA = 2a

AD ⊥ BC
D is mid point of BC
BD = DC = \(\frac { 1 }{ 2 }\) BC
= \(\frac { 1 }{ 2 }\) x 2a = a
Now, in right ∆ADB,
AB² = AD² + BD² (Pythagoras Theorem)
(2a)² = AD² + a²
⇒ 4a² – a² = AD²
⇒ AD² = 3a² = (√3 a)²
AD = √3 a = a√3 units

Question 12.
Solution:
∆ABC is an equilateral triangle in which
AB = BC = CA = 12 cm

AD ⊥ BC which bisects BC at D
BD = DC = \(\frac { 1 }{ 2 }\) BC = \(\frac { 1 }{ 2 }\) x 12 = 6cm
Now, in right ∆ADB,
AB² = AD² + BD²
⇒ (12)² = AD² + (6)²
⇒ 144 = AD² + 36
AD² = 144 – 36 = 108
AD = √108 = √(36 x 3) = 6√3 cm

Question 13.
Solution:
Let ABCD is a rectangle in which adjacent sides.
AB = 30 cm and BC = 16 cm

AC is its diagonal.
In right ∆ABC,
AC² = AB² + BC² (Pythagoras Theorem)
= (30)² + (16)² = 900 + 256 = 1156 = (34)²
Diagonal AC = 34 cm

Question 14.
Solution:
ABCD is a rhombus
Its diagonals AC and BD bisect each other at O.

AO = OC = \(\frac { 1 }{ 2 }\) AC.
and BO = OD = \(\frac { 1 }{ 2 }\) BD
BD = 24 cm and AC = 10 cm
BO = \(\frac { 1 }{ 2 }\) x BD = \(\frac { 1 }{ 2 }\) x 24 = 12 cm
AO = \(\frac { 1 }{ 2 }\) x AC = \(\frac { 1 }{ 2 }\) x 10 = 5 cm
Now, in right ∆AOB,
AB² = AO² + BO² = (5)² + (12)² = 144 + 25 = 169 = (13)²
AB = 13
Hence, each side of rhombus = 13 cm

Question 15.
Solution:
Given : In ∆ABC, AC > AB.
D is the mid point of BC and AE ⊥ BC.
AD is joined.

To prove: AB² = AD² – BC x DE + \(\frac { 1 }{ 4 }\) BC2
Proof: In ∆AEB, ∠E = 90°
AB² = AE² + BE² …..(i) (Pythagoras Theorem)
In ∆AED, ∠E = 90°
AD² = AE² + DE²
⇒ AE² = AD² – DE² …..(ii)
Now, substitute eq. (ii) in eq. (i)
AB² = AE² + BE²
AB² = AD² – DE² + BE² [from (ii)]
AB² = (AD² – DE²) + (BD – DE)² [BE = BD – DE²]

Question 16.
Solution:

Question 17.
Solution:
Given : In ∆ABC, D is the mid point of BC, AE ⊥ BC,
BC = a, AC = b, AB = c, ED = x, AD =p and AE =
AD is joined.
To prove :

Question 18.
Solution:
Given : In ∆ABC, AB =AC
BC is produced to D and AD is joined.
To prove : (AD² – AC²) = BD x CD
Construction : Draw AE ⊥ BC.

Proof: In ∆ABC,
AB = AC and AE ⊥ BC
BE = EC
Now, in right ∆AED, ∠E = 90°
AD² = AE² + ED² …..(i)
and in right ∆AEC, ∠E = 90°
AC² = AE² + EC² …..(ii)
Now, subtracting (i) and (ii),
AD² – AC² = (AE² + ED²) – (AE² + EC²)
= AE² + ED² – AE² – EC²
= ED² – EC²
= (ED + EC) (ED – EC) (BE = EC proved above)
= BD x CD = BD x CD
AD² – AC² = BD x CD
Hence proved.

Question 19.
Solution:
Given : In ∆ABC, AB = BC and ∠ABC = 90°
∆ACD and ∆ABE are similar to each other.

To prove : Ratio between area ∆ABE and area ∆ACD.
Proof: Let AB = BC = x
Now, in right ∆ABC,
⇒ AC² = AB² + BC² = AB² + AB² = 2AB² = 2x²
∆ABE and ∆ACD are similar

Ratio between the areas of ∆ABE and ∆ACD = 1 : 2

Question 20.
Solution:
An aeroplane flies from airport to north at the speed of 1000 km/hr.
Another aeroplane flies from the airport to west at the speed of 1200 km/hr.
Period = 1\(\frac { 1 }{ 2 }\) hours
Distance covered by the first plane in 1\(\frac { 1 }{ 2 }\) hours = 1000 x \(\frac { 3 }{ 2 }\) km = 1500 km
and distance covered by another plane in 1\(\frac { 1 }{ 2 }\) hours = 1200 x \(\frac { 3 }{ 2 }\) km = 1800 km
At present, the distance between them
AB² = (BO)² + (AO)²
= (1800)² + (1500)²
= 3240000 + 2250000
= 5490000

Question 21.
Solution:
Given : In ∆ABC,
D is the mid point of BC and AL ⊥ BC
AD is joined.
To prove:

Question 22.
Solution:
AM is rod and BC is string out of rod.
In ∆BMC,
BC² = BM² + CM² = (1.8)² + (2.4)²

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles are helpful to complete your math homework.

If you have any doubts, please comment below. A Plus Topper try to provide online math tutoring for you.

RS Aggarwal Solutions Class 10 Chapter 4 Triangles

These Solutions are part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 4 Triangles

Exercise 4E

Very-Short-Answer and Short-Answer Questions
Question 1.
Solution:
Two properties for similarity of two triangles are:
(i) Angle-Angle-Angle (AAA) property.
(ii) Angle-Side-Angle (ASA) property.

Question 2.
Solution:
In a triangle, if a line parallel to one side is drawn, it will divide the other two sides proportionally.

Question 3.
Solution:
If a line divides any two sides of a triangle in the same ratio. Then, the line must be parallel to the third side.

Question 4.
Solution:
The line joining the midpoints of two sides of a triangle, is parallel to the third side.

Question 5.
Solution:
In two triangles, if three angles of the one triangle are equal to the three angles of the other, the triangles are similar.

Question 6.
Solution:
In two triangles, if two angles of the one triangle are equal to the corresponding angles of the other triangle, then the triangles are similar.

Question 7.
Solution:
In two triangles, if three sides of the one are proportional to the corresponding sides of the other, the triangles are similar.

Question 8.
Solution:
In two triangles, if two sides of the one are proportional to the corresponding sides of the other and their included angles are equal, the two triangles are similar.

Question 9.
Solution:
In a right angled triangle, the square on the hypotenuse is equal to the sum of squares on the other two sides.

Question 10.
Solution:
In a triangle, if the square on the longest side is equal to the sum of the squares on the other two sides, then the angle opposite to the hypotenuse is a right angle.

Question 11.
Solution:
The ratio of their areas will be 1 : 4.

Question 12.
Solution:
In two triangles ∆ABC and ∆PQR,
AB = 3 cm, AC = 6 cm, ∠A = 70°
PR = 9 cm, ∠P = 70° and PQ= 4.5 cm

Question 13.
Solution:
∆ABC ~ ∆DEF
2AB = DE, BC = 6 cm

Question 14.
Solution:
In the given figure,
DE || BC
AD = x cm, DB = (3x + 4) cm
AE = (x + 3) cm and EC = (3x + 19) cm

Question 15.
Solution:
AB is the ladder and A is window.
AB =10 m, AC = 8 m

We have to find the distance of BC
Let BC = x m
In right ∆ABC,
AB² = AC² + BC² (Pythagoras Theorem)
(10)² = 8² + x²
⇒ 100 = 64 + x²
⇒ x² = 100 – 64 = 36 = (6)²
x = 6
Distance between foot of ladder and base of the wall = 6 m.

Question 16.
Solution:
∆ABC is an equilateral triangle with side = 2a cm
AD ⊥ BC
and AD bisects BC at D

Now, in right ∆ABD,
AB² = AD² + BD² (Pythagoras Theorem)
⇒ (2a)² = AD² + (a)²
⇒ 4a² = AD² + a²
⇒ AD² = 4a² – a² = 3a²
AD = √3 a² = √3 a cm

Question 17.
Solution:
Given : ∆ABC ~ ∆DEF
and ar (∆ABC) = 64 cm²
and ar (∆DEF) = 169 cm², BC = 4 cm.

Question 18.
Solution:
In trapezium ABCD,
AB || CD
AB = 2CD
Diagonals AC and BD intersect each other at O
and area(∆AOB) = 84 cm².

Question 19.
Solution:
Let ∆ABC ~ ∆DEF

Question 20.
Solution:
In an equiangular ∆ABC,
AB = BC = CA = a cm.
Draw AD ⊥ BC which bisects BC at D.

Question 21.
Solution:
ABCD is a rhombus whose sides are equal and diagonals AC and BD bisect each other at right angles.

∠AOB = 90° and AO = OC, BO = OD
AO = \(\frac { 24 }{ 2 }\) = 12 cm
and BO = \(\frac { 10 }{ 2 }\) = 5 cm
Now, in right ∆AOB,
AB² = AO² + BO² (Pythagoras Theorem)
= (12)² + (5)² = 144 + 25 = 169 = (13)²
AB = 13
Each side of rhombus = 13 cm

Question 22.
Solution:
∆DEF ~ ∆GHK
∠D = ∠G = 48°
∠E = ∠H = 57°
∠F = ∠K

Now, in ∆DEF,
∠D + ∠E + ∠F = 180° (Angles of a triangle)
⇒ 48° + 57° + ∠F = 180°
⇒ 105° + ∠F= 180°
⇒ ∠F= 180°- 105° = 75°

Question 23.
Solution:
In the given figure,
In ∆ABC,
MN || BC
AM : MB = 1 : 2

Question 24.
Solution:
In ∆BMP,
PB = 5 cm, MP = 6 cm and BM = 9 cm
and in ∆CNR, NR = 9 cm

Question 25.
Solution:
In ∆ABC,
AB = AC = 25 cm
BC = 14 cm

AD ⊥ BC which bisects the base BC at D.
BD = DC = \(\frac { 14 }{ 2 }\) = 7 cm
Now, in right ∆ABD,
AB² = AD² + BD² (Pythagoras Theorem)
(25)² = AD² + 7²
625 = AD² + 49
⇒ AD² = 625 – 49 = 576
⇒ AD² = 576 = (24)²
AD = 24 cm
Length of altitude = 24 cm

Question 26.
Solution:
A man goes 12 m due north of point O reaching A and then 37 m due west reaching B.

Join OB,
In right ∆OAB,
OB² = OA² + AB² (Pythagoras Theorem)
= (12)² + (35)² = 144 + 1225 = 1369 = (37)²
OB = 37 m
The man is 37 m away from his starting point.

Question 27.
Solution:
In ∆ABC, AD is the bisector of ∠A which meets BC at D.
AB = c, BC = a, AC = b

Question 28.
Solution:
In the given figure, ∠AMN = ∠MBC = 76°
p, q and r are the lengths of AM, MB and BC Express the length of MN in terms of p, q and r.
In ∆ABC,
∠AMN = ∠MBC = 76°
But there are corresponding angles
MN || BC
∆AMN ~ ∆ABC

Question 29.
Solution:
In rhombus ABCD,
Diagonals AC and BD bisect each other at O at right angles.

AO = OC = \(\frac { 40 }{ 2 }\) = 20 cm
BO = OD = \(\frac { 42 }{ 2 }\) = 21 cm
Now, in right ∆AOB,
AB2 = AO2 + BO2 = (20)2 + (21)2 = 400 + 441 = 841
AB = √841 cm = 29 cm

Each side of rhombus = 29 cm

For each of the following statements state whether true (T) or false (F):
Question 30.
Solution:
(i) True.
(ii) False, as sides will not be proportion in each case.
(iii) False, as corresponding sides are proportional, not necessarily equal.
(iv) True.
(v) False, in ∆ABC,
AB = 6 cm, ∠A = 45° and AC = 8 cm

(vi) False, the polygon joining the midpoints of a quadrilateral is not a rhombus but it is a parallelogram.
(vii) True.
(viii) True.
(ix) True, O is any point in rectangle ABCD then
OA² + OC² = OB² + OD² is true.

(x) True.

Hope given RS Aggarwal Solutions Class 10 Chapter 4 Triangles are helpful to complete your math homework.

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