Question

In triangle PQR, ∠Q = ∠R = 45° and QR = 10 cm. Find its area.

Answer 1

Given:

• Triangle PQR with ∠Q = ∠R = 45°
• Side QR = 10 cm

Step wise calculation:

1. Since ∠Q = ∠R = 45°, this is an isosceles right triangle at Q and R.

2. Therefore, sides PQ and PR are equal.

3. Using the property of isosceles right triangles, the side opposite the right angle here QR can be related to the other sides as follows: `QR = PQ sqrt(2)`

4. Let the equal sides PQ = PR = x. 

So, `xsqrt(2) = 10`

⇒ `x = 10/sqrt(2)`

⇒ `x = (10sqrt(2))/2`

⇒ `x = 5sqrt(2)  cm`

5. Area of triangle PQR can be calculated by: 

Area = `1/2 xx PQ xx PR xx sin 90^circ`

6. Since ∠P = 90° as sum of all angles = 180° and given ∠Q = ∠R = 45°, so ∠P = 90°: 

`"Area" = 1/2 xx x xx x`

= `1/2 xx (5sqrt(2))^2`

= `1/2 xx 25 xx 2`

= `1/2 xx 50`

= 25 cm²

The area of triangle PQR is 25 cm².