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

NCERT Solutions Class 7 Mathematics 
Chapter – 6 (The Triangle and its Properties)

The NCERT Solutions in English Language for Class 7 Mathematics Chapter – 6 The Triangle and its Properties Exercise 6.5 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 6 The Triangles and its Properties

• NCERT Solution Class 7 Maths Exercise – 6.1
• NCERT Solution Class 7 Maths Exercise – 6.2
• NCERT Solution Class 7 Maths Exercise – 6.3
• NCERT Solution Class 7 Maths Exercise – 6.4

Exercise – 6.5 

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

Solution –
Let us draw a rough sketch of right-angled triangle

In right angled triangle PQR, we have
By the rule of Pythagoras Theorem,
QR² = PQ² + PR²
⇒ (10)² + (24)²
⇒ 100 + 576 = 676
∴ QR = √676 = 26 cm
The, the required length of QR = 26 cm.

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

Solution –
Let us draw a rough sketch of right-angled triangle

In right angled ∆ABC, we have
By the rule of Pythagoras Theorem,
BC² + (7)² = (25)²
⇒ BC² + 49 = 625
⇒ BC² = 625 – 49
⇒ BC² = 576
∴ BC = √576 = 24 cm
Thus, the required length of BC = 24 cm.

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.
By the rule of Pythagoras Theorem,
∴ a² + (12)² = (15)²
⇒ a²+ 144 = 225
⇒ a2 = 225 – 144
⇒ a² = 81
∴ a = √ 81 = 9 m
Thus, the distance of the foot from the ladder = 9m

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, 2cm, 2.5 cm.
In the case of right-angled triangles, identify the right angles.

Solution –

(i) 2.5 cm, 6.5 cm, 6 cm.
Square of the longer side = (6.5)² = 42.25 cm.
Sum of the square of 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 other two sides.
∴ The given sides form a right triangle.

(ii) 2 cm, 2 cm, 5 cm.
Square of the longer side = (5)² = 25 cm
Sum of the square of 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) 1.5 cm, 2 cm, 2.5 cm
Square of the longer side = (2.5)² = 6.25 cm
Sum of the square of other two sides
⇒ (1.5)² + (2)² 
⇒ 2.25 + 4
⇒ 6.25 cm
Since the square of longer side in a triangle is equal to the sum of square of other two sides.
∴ The given sides form a right triangle.

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 ABC is the triangle and B is the point where tree is broken at the height 5 m from the ground.
Tree top touches the ground at a distance of AC = 12 m from the base of the tree,

By observing the figure we came to conclude that right angle triangle is formed at A.
From the rule of Pythagoras theorem,
BC² = AB² + AC²
BC² = 5² + 12²
BC² = 25 + 144
BC² = 169
BC = √169
BC = 13 m
Then, the original height of the tree = AB + BC
= 5 + 13
= 18 m

6. Angles Q and R of a ΔPQR are 25^o and 65^o. 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°
∆PQR is a right triangle, right angled at P

(i) PQ² + QR² = RP²
Not True
∴ PQ² + QR² ≠ RP² (By Pythagoras property)

(ii) PQ² + RP² = QR²
True
∴ PQ² + RP² = QP² (By Pythagoras property)

(iii) RP² + QR² = PQ²
Not True
∴ RP² + QR² ≠ PQ² (By Pythagoras property)

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

Solution –

Let ABCD be the rectangular plot.
Then, AB = 40 cm and AC = 41 cm
BC =?
According to Pythagoras theorem,
From right angle triangle ABC, we have:
⇒ AC² = AB² + BC²
⇒ 41² = 40² + BC²
⇒ BC² = 41² – 40²
⇒ BC² = 1681 – 1600
⇒ BC² = 81
⇒ BC = √81
⇒ BC = 9 cm
Hence, the perimeter of the rectangle plot = 2 (length + breadth)
Where, length = 40 cm, breadth = 9 cm
Then,
= 2(40 + 9)
= 2 × 49
= 98 cm

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°.
∴ OA = OC = 8 cm and OB = OD = 15 cm
In right ∆OAB,
According to Pythagoras theorem,
⇒ AB² = OA² + OB²
⇒ AB² = (8)²+ (15)² 
⇒ AB² = 64 + 225
⇒ AB² = 289
⇒ AB = √289
⇒ AB = 17 cm
Since AB = BC = CD = DA (Property of rhombus)
∴ Required perimeter of rhombus
= 4 × side = 4 × 17 = 68 cm.