RBSE Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5

Rajasthan Board RBSE Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 Textbook Exercise Questions and Answers.

RBSE Class 7 Maths Solutions Chapter 6 The Triangles and Its Properties Ex 6.5

Question 1.
PQR is a triangle, right-angled at P. If PQ = 10 cm and PR = 24 cm, find QR.

Answer:
QR² = PQ² + PR² (By Pythagoras property)
⇒ QR² = 10² + 24²
⇒ QR² = 100 + 576 = 676
∴ QR = 26 cm

Question 2.
ABC is triangle, right-angled at C. If AB = 25 cm and AC = 7 cm, find BC.

Answer:
AC² + BC² = AB² (By Pythagoras property)
⇒ 7² + BC² = 25²
⇒ BC² = 625 - 49
⇒ BC² = 576
∴ BC = 24 cm

Question 3.
A 15 m long ladder reached a window 12 m high from the ground on placing it against a wall at a distance a. Find the distance of the foot of the ladder from the wall.

Answer:
Let the distance of the foot-of the from the wall is a m. Then
a² + 12² = 15² (By Pythagoras property)
a² = 225 - 144
a² = 81 ∴ a = 9m
∵ The distance of the foot of the ladder from the wall is 9 m.

Question 4.
Which of the following can be the sides of a right triangle?
(i) 2.5 cm, 6.5 cm, 6 cm
(ii) 2 cm, 2 cm, 5 cm
(iii) 1.5 cm, 2 cm, 2.5 cm
In the cose of right-angled triangles, identify the right angles.
Answer:
(i) 2.5 cm, 6.5 cm and 6 cm
(2.5)² + 6² = 6.25 + 36 = 42.25
(6.5)² = 42.25
∴ (2.5)² + 6² = (6.5)²
∵ The given lengths can be the sides of a right triangle, the angle between the lengths 2.5 cm and 6 cm is a right angle.

(ii) 2 cm, 2 cm and 5 cm
(2)² + (2)² =4 + 4 = 8
5² = 25
∴ 2² + 2² ≠ 5²
∵ The given lengths cannot be the sides of the triangle.

(iii) 1.5 cm, 2 cm and 2.5 cm
(1.5)² + 2² = 2.25 + 4 = 6.25
(2.5)² = 6.25
∴ 2² + (1.5)² = (2.5)²
∵ The given length can be the sides of a right triangle, the angle between the lengths 2 cm and 1.5 cm is a right angle.

Question 5.
A tree is broken at a height of 5 m from the ground and its top touches the ground at a distance of 12 m from the base of the tree. Find the original height of the tree.

Answer:
Given that
AC = CD
In ΔCBD,
CB² + BD² = CD²
⇒ 12² + 5² = CD²
⇒ CD² = 144 + 25
⇒ CD² = 169
⇒ CD = 13
Hence, AC = 13 m
∴ AB = AC + CB = 13 + 5 = 18 m
∴ The original height of the tree is 18 m.

Question 6.
Angles Q and R of a ΔPQR are 25° and 65°. Write which of the following is true:
(i) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²
Answer:
We know that
∠P + ∠Q + ∠R = 180° (Angle sum property)
⇒ ∠P + 25° + 65° = 180°
⇒ ∠P = 180° - 90° = 90°
ΔPQR is a right triangle, right angled at P.

(ii) PQ² + RP² = QR² is true. (By Pythagoras property)

Question 7.
Find the perimeter of the rectangle whose length is 40 cm and diagonal is 41 cm.

Answer:
Let PQRS is a rectangle whose length is 40 cm.
In ΔQPS,
QP² + PS² = QS²
⇒ 40² + PS² = 41²
⇒ PS² = 41² - 40²
⇒ PS² = 1681 - 1600
⇒ PS² = 81
∴ PS = 9 cm
Now, the perimeter of the triangle is :
2 (PS + PQ) = 2(9 + 40)
= 2 × 49 = 98 cm.

Question 8.
The diagonals of a rhombus measure 16 cm and 30 cm. Find its perimeter.

Answer:
Let ABCD is a rhombus whose diagonals AC and BD are of length 16 cm and 30 cm respectively. Let the diagonals intersect at O. Since the diagonals of a rhombus bisect each other at right angles.
BO = OD = 15 cm
AO = OC = 8 cm, and
∠AOB = ∠DOC = ∠BOC = ∠AOD = 90°

In right-angled triangle AOB,
AO² + OB²=AB²
⇒ AB² = 15² + 8² = 225 + 64
⇒ AB² = 289 ⇒ AB = 17 cm
∴ Perimeter of a rhombus = 4 × side = 4 × 17 cm = 68 cm