## Chapter 6 Triangle 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.**

**Answer:**

PR = leg 1 = 24 cm

PQ = leg 2 = 10 cm

QR = hypotenuse = ?

By Pythagoras property

= (hyp)² = (leg 1)² + (leg 2)²

= x² = 24² + 10²

= x² = 576 + 100

= x = √676

= x = 26 cm

Therefore, by Pythagoras property QR is 26 cm.

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

AC = leg 1 = 7 cm

CB = leg 2 = ?

AB = hypotenuse = 25 cm

By Pythagoras property

= (hyp)² = (leg 1)² + (leg 2)²

= 25² = 7² + x²

= 625 = 49 + x²

= 625 – 49 = x²

= √576 = x

= 24 cm = x

Therefore, by Pythagoras property BC is 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:

Answer:

12 m = leg 1 = 12 m

a = leg 2 = ?

15 m = hypotenuse = 15 m

(by Pythagoras property)

= (hyp)² = (leg 1)² + (leg 2)²

= 15² = 12² + x²

= 225 = 144 + x²

= 225 – 144

= x = 81

= √81

= 9 m

Therefore, by Pythagoras property, 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, 2cm, 2.5 cm

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

Answer:

(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.

Answer:

**(i)**Let a = 2.5 cm = leg 1

b = 6 cm = leg 2

c = 6.5 cm = hypotenuse

(by Pythagoras property)

= (hyp)² = (leg 1)² + (leg 2)²

= 6.5² = 2.5² + 6²

= 42.25 = 6.25 + 36

Therefore, by the property with these measurements we can make a triangle. The right angle is opposite to hypotenuse of the triangle.

**(ii)**Let a = 2 cm = leg 1

b = 2 cm = leg 2

c = 5 cm = hypotenuse

(by Pythagoras property)

= (hyp)² = (leg 1)² + (leg 2)²

= 5² = 2² + 2²

= 25 = 4 + 4

Therefore, by the property with these measurements we cannot make a triangle.

**(iii)**Let a = 1.5 cm = leg 1

b = 2 cm = leg 2

c = 2.5 cm = hypotenuse

(by Pythagoras property)

= (hyp)² = (leg 1)² + (leg 2)²

= 2.5² = 1.5² + 2²

= 6.25 = 2.25 + 4

Therefore, by the property with these measurements we can make a triangle. The right angle is opposite to hypotenuse of the triangle.

**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:

Answer:

AB = leg 1 = 5 m

AC = leg 2 = 12 m

BC = hypotenuse = x

(by Pythagoras property)

In ΔABC,

= (hyp)² = (leg 1)² + (leg 2)²

= x² = 5² + 12²

= x² = 25 + 144

= x² = 169

= x² = √169 = 13 m

Therefore, BC is 13 m.

Original height of the tree = AB + BC

= 5 + 13

= 18 m

Therefore, the original height of tree is 18 m.

**Question 6: Angles Q and R of a ΔPQR are 25o and 65o.**

Write which of the following is true:

(i) PQ2 + QR2 = RP2

(ii) PQ2 + RP2 = QR2

(iii) RP2 + QR2 = PQ2

Answer:

Write which of the following is true:

(i) PQ2 + QR2 = RP2

(ii) PQ2 + RP2 = QR2

(iii) RP2 + QR2 = PQ2

Answer:

(by angle sum property)

The sum of measure of angles = 180°

= 180° = x + 25° + 65°

= 180 - (25 + 65)

= 180 – 90 = 90°

Therefore, by the property angle P = 90°. The right angle is opposite to hypotenuse. So, QR is the hypotenuse.

PQ = leg 1

PR = leg 2

QR = hypotenuse

(by Pythagoras property)

= (hyp)² = (leg 1)² + (leg 2)²

= QR² = PQ² + PR²

Therefore, option 2 is correct.

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

Answer:

Answer:

AC = hypotenuse = 41 cm

BC = leg 1 = ?

AB = leg 2 = 40 cm

(by Pythagoras property)

= (hyp)² = (leg 1)² + (leg 2)²

= 41² = x² + 40²

= 1681 = x² + 1600

= 1681 – 1600 = 81

= √81 = 9 cm

Therefore, BC is 9 cm.

Length of rectangle = 40 cm

Breadth of rectangle = 9 cm

Perimeter of rectangle = 2 x (l + b)

= 2 x (40 + 9)

= 2 x 49

= 98 cm

Therefore, the perimeter of rectangle is 98 cm.

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

Answer:

Answer:

Length of AC = 16 cm

Length of BD = 30 cm

△AOD is a right angled triangle.

Let AD = x

(by Pythagoras property)

= (hyp)² = (leg 1)² + (leg 2)²

= x² = 8² + 15²

= x² = 64 + 225

= x² = 289 = √289

= x² = 17 cm

Perimeter of Rhombus = 4 x side

= 4 x 17

= 68 cm

Therefore, the perimeter of the rhombus is 68 cm.

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