These NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Ex 6.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 7 Maths Chapter 6 The Triangles and Its Properties Exercise 6.5

Question 1.

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

Answer:

In the right ΔPQR

QR^{2} = PQ^{2} + PR^{2}

(pythagoras property)

= 10^{2} + 24^{2}

= 100 + 576

= 676

QR^{2} = 26^{2}

∴ QR = 26 cm

Question 2.

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

Answer:

In the right ΔABC

AB^{2} = AC^{2} + BC^{2}

(using pythagoras property)

25^{2} = 7^{2} + BC^{2}

625 = 49 + BC^{2}

625 – 49 = BC^{2}

576 = BC^{2}

24^{2} = BC^{2}

∴ 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 a ladder from the wall be ‘a’m

a2^{2} + 12^{2} = 15^{2}

(using pythagoras property)

a^{2} + 144 = 225

a^{2} = 225 – 144

a^{2} = 81

a^{2} = 9^{2}

a = 9 m

The distance of the foot of the ladder from the wall = 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 case of right-angled triangles, identify the right angles.

Answer:

(i) 2.5 cm, 6.5 cm, 6 cm^{2}.

The longest side is 6.5 cm

2.5^{2} + 6^{2} = 6.25 + 36

= 42.25

6.5^{2} = 42.25

∴ 6.5^{2} = 2.5^{2} + 6^{2}

(pythagoras property)

The given lengths can be the sides of a right triangle.

The right angle is the angle between the sides 2.5 cm and 6 cm

(ii) 2 cm, 2 cm, 5 cm

The longest side is 5 cm

2^{2} + 2^{2} = 4 + 4 = 8

5^{2} = 25

5^{2} ≠ 2^{2} + 2^{2}

∴ The given length cannot be the sides of a right triangle.

(iii) 1.5 cm, 2 cm, 2.5 cm

The longest side is 2.5 cm

1.5^{2} + 2^{2} = 2.25 + 4

= 6.25

2.5^{2} = 6.25

1.5^{2} + 2^{2} = 2.5^{2}

(pythagoras property)

The given lengths can be sides of a right triangle.

The right angle is the angle between the sides 1.5 cm and 2 cm.

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:

Let the tree BD be broken at the point C, such that CD = CA

In the right triangle ABC, using the pythagoras property; we get

AC^{2} = AB^{2} + BC^{2}

AC^{2} = 12^{2} + 5^{2}

= 144 + 25

AC^{2} = 169

AC^{2} = 13^{2}

∴ AC = 13

Now, height of the tree

BD = BC + CD (CD = AC)

= 5 m + 13 m

= 18 m

The height of the tree is 18 m.

Question 6.

Angles Q and R of PQR are 25° and 65°. Write which of the following is true:

(i) PQ^{2} + QR^{2} = RP^{2}

(ii) PQ^{2} + RP^{2} = QR^{2}

(iii) RP^{2} + QR^{2} = PQ^{2}

Answer:

In ΔPQR

∠P + ∠Q +∠R = 180°

∠P + 25° + 65° = 180°

∠P + 90° = 180°

∠P = 180° – 90°

= 90°

So, ΔPQR is a triangle right angled at P.

∴ QR is the hypotenuse.

∴ using the pythagoras property

QR^{2} = PQ^{2} + PR^{2} So,

(ii) PQ^{2} + RP^{2} = QR^{2} is true

Question 7.

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

Answer:

Let the breadth AD be x cm.

In the right triangle ABD,

BD^{2} = AD^{2} + AB^{2}

41^{2}= x^{2} + 40^{2}

x^{2} = 41^{2} – 40^{2}

= 1681 – 1600

x^{2} = 81

x^{2} = 9^{2}

x = 9 cm

Perimeter of the rectangle = 2 (l + b)

= 2 (40 + 9)

= 2 × 49

= 98 cm

Thus, the perimeter of the rectangle

= 98 cm

Question 8.

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

Answer:

Since the diagonals of a rhombus bisect each other at right angles.

∠AOB = ∠BOC = ∠COD = ∠AOD = 90°

AO = \(\frac { 1 }{ 2 }\) AC; AO = \(\frac { 1 }{ 2 }\) × 30 = 15

∵ AO = OC = 15 cm and BO = OD = 8 cm

In the right ΔAOB,

AB^{2} = AO^{2} + BO^{2}

AB^{2} = 15^{2} + 8^{2}

AB^{2} = 225 + 64

AB^{2} = 289

AB^{2} = 17^{2}

AB = 17 cm

∵ Side of the rhombus is 17 cm.

∴ Perimeter of the rhombus ABCD = 4 × 17

(all the four sides are equal) = 68 cm