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 diagram of right-angled triangle:

In the above figure, RQ is the hypotenuse, PR is the base and PQ is the height

QR² = PQ² + PR²  (From the rule of Pythagoras property)

QR² = 10² + 24²

QR² = 100 + 576

QR² = 676

QR = √676

QR = 26 cm

Hence, the length of the hypotenuse 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 diagram of right-angled triangle:

In the above figure, RQ is the hypotenuse,

AB² = AC² + BC²  (From the rule of Pythagoras Property)

252 = 72 + BC²

625 = 49 + BC²

Transposing 49 from RHS to LHS it becomes – 49

BC² = 625 – 49

BC² = 576

BC = √576

BC = 24 cm

Hence, the length of the 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:

In the above figure RQ is the hypotenuse,

15² = 12² + a² (From the rule of Pythagoras Property)

225 = 144 + a²

Transposing 144 from RHS to LHS it becomes – 144

a² = 225 – 144

a² = 81

a = √81

a = 9 m

Hence, the length of a = 9 m.

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) Let a = 2.5 cm, b = 6.5 cm, c = 6 cm

Let the largest value is the hypotenuse of the right-angled triangle i.e. b = 6.5 cm.

According to the rule of the Pythagoras theorem, we know that the square of the longer side is equal to the sum of the square of the other two sides.

b² = a² + c²

(6.5)² = (2.5)² + 6²

42.25 = 6.25 + 36

42.25 = 42.25

Here the sum of the square of two sides of the triangle is equal to the square of the third side,

Hence the given triangle is a right-angled triangle.

The right angle lies on the opposite of the greater side 6.5 cm.

(ii) Let a = 2 cm, b = 2 cm, c = 5 cm

Let the largest value is the hypotenuse side of the right-angled triangle i.e. c = 5 cm.

C² = a²+ b² (From the rule of Pythagoras Property)

5² = 2² + 2²

25 = 4 + 4

25 ≠ 8

Here, the sum of the square of the two sides of the triangle is not equal to the square of the third side.

Hence, the given triangle is not a right-angled triangle.

(iii) Let a = 1.5 cm, b = 2 cm, c = 2.5 cm

Let the largest value is the hypotenuse side of right-angled triangle i.e. b = 2.5 cm.

b² = a² + c² (From the rule of Pythagoras Property)

(2.5)² = (1.5)² + 2²

6.25 = 2.25 + 4

6.25 = 6.25

Here, the sum of the square of the two sides of the triangle is equal to the square of the third side.

Hence, the given triangle is a right-angled triangle.

The right angle lies on the opposite of the greater side 2.5 cm.

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 PQR is the triangle and Q is the point where tree is broken at the height 5 m from the ground.

Tree top touches the ground at a distance of PR = 12 m from the base of the tree.

By observing the figure we came to conclude that right angle triangle is formed at P.

QR² = PQ² + PR² (From the rule of Pythagoras theorem)

QR² = 5² + (12)²

QR² = 25 + 144

QR²= 169

QR = √169

QR = 13 m

Then, the original height of the tree = PQ + QR

= 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:

Given,

 ∠Q = 25^o, ∠R = 65^o

Then we have to find,

∠P =?

We know that sum of the three interior angles of triangle is equal to 180^o.

So,

∠PQR + ∠QRP + ∠RPQ = 180^o

25^o + 65^o + ∠RPQ = 180^o

90^o + ∠RPQ = 180^o

∠RPQ = 180 – 90

∠RPQ = 90^o

We also know that the side opposite to the right angle is the hypotenuse of that right-angled triangle.

So, QR² = PQ² + PR² Hence, (ii) is true

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

Solution:

First, we draw a rough sketch according to the question:

Let ABCD be the rectangle.

Given, AB = 40 cm and AC = 41 cm

BC =?

According to the rule of Pythagoras theorem,

From the right-angled triangle ABC,

AC² = AB² + BC²

41² = 40² + BC²

BC² = 41² – 40²

BC² = 1681 – 1600

BC² = 81

BC = √81

BC = 9 cm

length = 40 cm, breadth = 9 cm

Now, the perimeter of the rectangle = 2 (length + breadth)

Then,

= 2(40 + 9)

= 2 × 49

= 98 cm

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

Solution:

First, we draw a rough sketch according to the question:

Let PQRS be a rhombus, all sides of the rhombus have equal length and its diagonal PR and SQ are intersecting each other at a point O. Diagonals in the rhombus bisect each other at 90^o.

So, PO = (PR/2)

= 16/2

= 8 cm

And, SO = (SQ/2)

= 30/2

= 15 cm

Then we consider the triangle POS and apply the Pythagoras theorem,

PS² = PO² + SO²

PS² = 82 + 152

PS² = 64 + 225

PS² = 289

PS = √289

PS = 17 cm

Hence, the length of side of rhombus is 17 cm

Now,

Perimeter of rhombus = 4 × side of the rhombus

= 4 × 17

= 68 cm

 Hence, Perimeter of rhombus is 68 cm.