## RS Aggarwal Class 7 Solutions Chapter 15 Properties of Triangles Ex 15D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15D.

Other Exercises

Question 1.
Solution:
In right triangle ABC, ∠B = 90° AB = 9cm, BC = 12cm By Pythagoras Theorem,
AC² = AB² + BC² = (9)² + (12)² = 81 + 144 = 225
AC = √225 = 15 cm

Question 2.
Solution:
In right ∆ABC, ∠B = 90°
AC = 26cm, AB = 10cm
By Pythagoras Theorem
AC² = AB² + BC² ⇒ (26)² = (10)² + BC²
⇒ 676 = 100 + BC²
⇒ BC² = 676 – 100 = 576 = (24)²
⇒ BC = 24 cm

Question 3.
Solution:
In right ∆ABC, ∠C = 90°,
AB = 7.5cm, BC = 4.5cm
By Pythagoras Theorem
AB² = BC² + AC² ⇒ (7.5)² = (4.5)² + AC²
⇒ 56.25 = 20.25 + AC²
⇒ AC² = 56.25 – 20.25 = 36.00 = (6)²
⇒ AC = 6cm

Question 4.
Solution:
In ∆ABC, ∠B = 90°
Let each leg = x cm By Pythagoras Theorem,
x² + x² = AC²
⇒ 2x² = 50
⇒ x² = 25 = (5)²
⇒ x = 5
Length of each equal leg = 5cm

Question 5.
Solution:
A triangle is a right-angled,
If (Hypotenuse)² = sum of squares or other two sides
If (39)² = (15)² + (36)² (Hypotenuse is the longest side)
If 1521 = 225 + 1296
If 1521 = 1521 Which is true.
It is a right-angled triangle.

Question 6.
Solution:
In ∆ABC, ∠C = 90° a = 6cm, b = 4.5cm.
By Pythagoras Theorem
c² = a² + b² = (6)² + (4.5)² = 36.00 + 20.25 = 56.25 = (7.5)²
c = 7.5 cm

Question 7.
Solution:
A triangle will be a right angled
if (longest side)² = Sum of squares of other two sides
(i) a = 15cm, b = 20cm, c = 25cm.
Here, longest side = c ,
The triangle will be right angled
if c² = a² + b²
if (25)² = (15)² + (20)²
if 625 = 225 + 400 = 625 Which is true.
It is a right angled triangle.
(ii) a = 9cm, b = 12cm, c = 16cm
∆ABC is a right angled triangle if
c² = a² + b²
if (16)² = (9)² + (12)²
if 256 = 81 + 144 = 225
⇒ 256 = 225
Which is not true
Triangle is not a right angled triangle.
(iii) a = 10cm, b = 24cm, c = 26cm
The triangle ABC is a right angled triangle
if c² = a² + b²
if (26)² = (10)² + (24)²
if 676 = 100 + 576
if 676 = 676 Which is true.
The triangle is a right angled triangle.

Question 8.
Solution:
In ∆ABC,
∠B = 35° and ∠C = 55°
∠A = 180°- (∠B + ∠C) = 180° – (35° + 55°) = 180° – 90° = 90°
∆ABC is a right angled triangle
By Pythagoras Theorem,
BC² = AB² + AC²
(iii) is hue

Question 9.
Solution:
AB is a ladder and it is 15 m long B is window and BC = 12 m
In right ∆ABC
AB² = AC² + BC² (By Pythagoras Theorem)
⇒ (15)² = x² + (12)²
⇒ (15)² = x² + (12)²
⇒ 225 = x² + 144
⇒ x² = 225 – 144
⇒ x² = 81 = (9)²
x = 9 m
Distance of the foot of ladder from the wall = 9 m

Question 10.
Solution:
Let AB be the ladder and AC be the height. Length of ladder AB = 5m
and height CA = 4.8m
Let distance of the ladder from the wall BC = x
Now in right angled ∆ABC, ∠C = 90°
AB² = AC² + BC² (By Pythagoras Theorem)
⇒ (5)² = (4.8)² + x²
⇒ 25 = 23.04 + x²
⇒ x² = 25.00 – 23.04 = 1.96 = (1.4)²
⇒ x = 1.4
The foot of ladder are 1.4m away from the wall.

Question 11.
Solution:
Let AB be the tree which broke at D and its top A touches the ground at C
their BD = 5m, BC = 12m,
Let AD = x m, then CD = x m
Now, in right ∆ABC,
CD² = BD² + BC²
(By Pythagoras Theorem) CD² = (9)² + (12)² = 81 + 144 = 225 = (15)²
CD = 15m,
Height of the tree AB = AD + BD = 15 + 9 = 24m

Question 12.
Solution:
AB and CD are two poles and they are 12,m apart
AB = 18 m, CD = 13m and BD = 12 m From C, draw CE || BD Then
CE = BD = 12 m
and AE = AB – EB = AB – CD = 18 – 13 = 5 m
Join AC
Now in right ∆ACE
AC² = CE² + AE²
(By Pythagoras Theorem)
AC² = (12)² + (5)² = 144 + 25 = 169 = (13)²
AC = 13 m
Distance between their tops = 13 m

Question 13.
Solution:
A man starts from O and goes 35m due west and then 12m due north, then
In rights ∆OAB,
OA = 35 m
AB = 12 m OB² = OA² + AB² (By Pythagoras Theorem)
= (35)² + (12)² = 1225 + 144 = 1369 = (37)²
OB = 37
Hence he is 37m away from the starting point

Question 14.
Solution:
A man goes 3km due north and then 4km east. In right angled ∆OAB,
OA = 3km.
AB = 4km.
OB² = OA² + AB² (By Pythagoras Theorem)
= (3)² + (4)² = 9 + 16 = 25 = (5)²
OB = 5km
Hence he is 5km from the initial position.

Question 15.
Solution:
ABCD is a rectangle whose sides
AB = 16cm and BC = 12cm.
AC is its diagonal
In right angled ∆ABC
AC² = AB² + BC²
(By Pythagoras Theorem)
= (16)² + (12)² = 256 + 144 = 400 = (20)²
AC = 20cm
Hence length of diagonal AC = 20 cm

Question 16.
Solution:
ABCD is a rectangle and AC is its diagonal AB = 40 cm and AC = 41 cm
Now in right ∆ABC
AC² = AB² + BC² (By Pythagoras Theorem)
⇒ (41)² = (40)² + BC²
⇒ 1681 = 1600 + BC²
⇒ BC² = 1681 – 1600 = 81 = (9)²
⇒ BC = 9 cm
Now perimeter of rectangle ABCD = 2 (AB + BC)
= 2 (40 + 9) = 2 x 49 = 98 cm

Question 17.
Solution:
Perimeter of rhombus ABCD = 4 x Side
Diagonal AC = 30 cm and BD = 16 cm
The diagonals of rhombus bisect each other at right angles
AO = OC = $$\frac { 30 }{ 2 }$$ = 15 cm
and BO = OD = $$\frac { 16 }{ 2 }$$ = 8 m
Now in right ∆AOB,
AB² = AO² + BO² = (15)² + (8)² = 225 + 64 = 289 = (17)²
AB = 17 cm
Now perimeter = 4 x side = 4 x 17 = 68 cm

Question 18.
Solution:
(i) In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
(ii) If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right angled.
(iii) Of all the line segments that can be drawn to a given line from a given point outside it, the perpendicular is the shortest.

Hope given RS Aggarwal Solutions Class 7 Chapter 15 Properties of Triangles Ex 15D are helpful to complete your math homework.

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