NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 are the part of NCERT Solutions for Class 7 Maths (Rationalised Contents). Here you can find the NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5.

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.1
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.2
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.3
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.4
NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5

NCERT Solutions for Class 7 Maths Chapter 6 The Triangle and its Properties Ex 6.5 (Rationalised Contents)

Ex 6.5 Class 7 Maths Question 1.

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

Solution:
In right-angled triangle PQR, we have
QR² = PQ² + PR²               (Using Pythagoras property)
        = (10)² + (24)²
        = 100 + 576 = 676

∴ QR = √676 = 26 cm
Thus, the required length of QR is 26 cm.

Ex 6.5 Class 7 Maths Question 2.

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

Solution:
In right-angled ∆ABC, we have
BC² + AC² = AB² (Using Pythagoras property)
⇒ BC² + (7)² = (25)²

⇒ BC² + 49 = 625
⇒ BC² = 625 – 49
⇒ BC² = 576
∴ BC = √576 = 24 cm
Thus, the required length of BC is 24 cm.

Ex 6.5 Class 7 Maths 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.

Solution:
Here, the ladder forms a right-angled triangle.
∴ a² + (12)² = (15)² (Using Pythagoras property)
⇒ a²+ 144 = 225
⇒ a² = 225 – 144
⇒ a² = 81
∴ a = √81 = 9 m
Thus, the distance of the foot of the ladder from the wall is 9 m.

Ex 6.5 Class 7 Maths 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 case of right-angled triangles, identify the right angles.

Solution:
(i) Given sides are: 2.5 cm, 6.5 cm, 6 cm
Square of the longer side = (6.5)² = 42.25 cm
Sum of the squares of the other two sides
= (2.5)² + (6)²

= 6.25 + 36
= 42.25 cm
Since, the square of the longer side in a triangle is equal to the sum of the squares of the other two sides.
∴ The given sides form a right triangle. The angle opposite to the side 6.5 cm is right angle.

(ii) Given sides are: 2 cm, 2 cm, 5 cm
Square of the longer side = (5)² = 25 cm

Sum of the squares of the other two sides
= (2)² + (2)² = 4 + 4 = 8 cm
Since 25 cm ≠ 8 cm
∴ The given sides do not form a right triangle.

(iii) Given sides are: 1.5 cm, 2 cm, 2.5 cm
Square of the longer side = (2.5)² = 6.25 cm

Sum of the squares of the other two sides
= (1.5)² + (2)² = 2.25 + 4 = 6.25 cm
Since 6.25 cm = 6.25 cm
Since the square of the longer side in a triangle is equal to the sum of the squares of the other two sides.
∴ The given sides form a right triangle. The angle opposite to the side 2.5 cm is right angle.

Ex 6.5 Class 7 Maths 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.

Solution:

Let AB be the original height of the tree which is broken at C and touching the ground at D such that AC = 5 m and AD = 12 m
In right-angled triangle ∆CAD,
AD² + AC² = CD² (Using Pythagoras property)
⇒ (12)² + (5)² = CD²
⇒ 144 + 25 = CD²
⇒ 169 = CD²
∴ CD = √169 = 13 m
But BC = CD = 13 cm
The tree, AB = AC + CB
AB = 5 m + 13 m
∴ AB = 18 m
Thus, the original height of the tree is 18 m.

Ex 6.5 Class 7 Maths Question 6.

Angles Q and R of a ∆PQR are 25° and 65°. Write which of the following is true.
(i) PQ² + QR² = RP²
(ii) PQ² + RP² = QR²
(iii) RP² + QR² = PQ²

Solution:
We know that
∠P + ∠Q + ∠R = 180°         (Angle sum property)
∠P + 25° + 65° = 180°
∠P + 90° = 180°
∠P = 180° – 90° = 90°
Therefore, ∆PQR is a right triangle, right angled at P.
(i) False
∴ PQ² + QR² ≠ RP²         (Using Pythagoras property)
(ii) True
∴ PQ² + RP² = QP²         (Using Pythagoras property)
(iii) False
∴ RP² + QR² ≠ PQ²         (Using Pythagoras property)

Ex 6.5 Class 7 Maths Question 7.

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

Solution:
It is given that, Length AB = 40 cm
Diagonal AC = 41 cm
In right-angled triangle ABC, we have
AB² + BC²= AC² (Using Pythagoras property)
⇒ (40)² + BC² = (41)²
⇒ 1600 + BC² = 1681
⇒ BC² = 1681 – 1600
⇒ BC² = 81
∴ BC = √81 = 9 cm
∴ The required perimeter of the rectangle = 2(AB + BC)
= 2(40 + 9) cm
= 98 cm

Ex 6.5 Class 7 Maths Question 8.

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

Solution:
Let ABCD be a rhombus whose diagonals intersect each other at O such that AC = 16 cm and BD = 30 cm
Since, the diagonals of a rhombus bisect each other at 90°.
Therefore, OA = OC = 8 cm and OB = OD = 15 cm
In right-angled ∆OAB,
AB² = OA² + OB²       (Using Pythagoras property)
      = (8)²+ (15)² = 64 + 225
       = 289
∴ AB = √289 = 17 cm
Since AB = BC = CD = DA           (Using property of a rhombus)
∴ Required perimeter of the rhombus
= 4 × side = 4 × 17 = 68 cm