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

**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 From Pythagoras property)

= (10)2 + (24)2

= 100 + 576 = 676

∴ QR = = 26 cm

The, the required length of QR = 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:**

BC2 + (7)2 = (25)2 (By Pythagoras property)

⇒ BC2 + 49 = 625

⇒ BC2 = 625 – 49

⇒ BC2 = 576

∴ BC = = 24 cm

Thus, the required length of BC = 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:**

In right angled∆ABC, we have

Solution:

Here, the ladder forms a right angled triangle.

∴ a^{2} + (12)^{2} = (15)^{2} (By Pythagoras property)

⇒ a^{2}+ 144 = 225

⇒ a^{2} = 225 – 144

⇒ a^{2} = 81

∴ a = = 9 m

Thus, the distance of the foot from the ladder = 9m

**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**

**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 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) Given sides are 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) 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 square of other two sides

= (1.5)² + (2)²= 2.25 + 4

Since 6.25 cm = 6.25 cm = 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.

**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 and broken at C touching the ground at D such that

AC = 5 m

and AD = 12 m

In right triangle ∆CAD,

AD² + AC² = CD² (By Pythagoras property)

⇒ (12)² + (5)² = CD²

⇒ 144 + 25 = CD²

⇒ 169 = CD²

∴ CD = = 13 m

But CD = BC

AC + CB = AB

5 m + 13 m = AB

∴ AB = 18 m .

Thus, the original height of the tree = 18 m.

**Ex 6.5 Class 7 Maths Question 6.**

**Angles Q and R of a APQR 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°

∆PQR is a right triangle, right angled at P

(i) Not True

∴ PQ² + QR² ≠ RP² (By Pythagoras property)

(ii) True

∴ PQ² + RP² = QP² (By Pythagoras property)

(iii) Not True

∴ RP² + QR² ≠ PQ² (By 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:**

Given: Length AB = 40 cm

Diagonal AC = 41 cm

In right triangle ABC, we have

AB² + BC²= AC² (By Pythagoras property)

⇒ (40)² + BC² = (41)²

⇒ 1600 + BC² = 1681

⇒ BC2 = 1681 – 1600

⇒ BC2 = 81

∴ BC = = 9 cm

∴ AB = DC = 40 cm and BC = AD = 9 cm (Property of rectangle)

∴ The required perimeter

= AB + BC + CD + DA

= (40 + 9 + 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°.

∴ OA = OC = 8 cm and OB = OD = 15 cm

In right ∆OAB,

AB² = OA² + OB² (By Pythagoras property)

= (8)²+ (15)² = 64 + 225

= 289

∴ AB = = 17 cm

Since AB = BC = CD = DA (Property of rhombus)

∴ Required perimeter of rhombus

= 4 × side = 4 × 17 = 68 cm.