Question

In trapezium PQRS, PQ = 20 cm, PR = 61 cm, SR = 60 cm and PS ⊥ SR. Find its area.

Answer 1

Given:

• In trapezium PQRS,
• PQ = 20 cm
• PR = 61 cm   ...(Diagonal)
• SR = 60 cm
• PS ⊥ SR   ...(PS is perpendicular to SR)

Stepwise calculation:

1. Since PS is perpendicular to SR, PS is the height of the trapezium when SR is taken as the base.

2. Use Pythagoras theorem in right triangle PSR to find PS:

PR serves as hypotenuse, SR and PS are perpendicular sides.

PR² = PS² + SR²

Given PR = 61 cm, SR = 60 cm,

So calculate PS:

61² = PS² + 60² 

3721 = PS² + 3600 

PS² = 3721 – 3600 = 121

PS = `\sqrt(121)`

PS = 11 cm

3. Now, find the length QR other parallel side by subtracting PQ from PR’s horizontal projection.

Since PS is the height, project PQ and QR on the same line SR to get their bases.

4. The area of trapezium is given by:

Area = `1/2` × (PQ + SR) × PS

Using values:

= `1/2 xx (20 + 60) xx 11`

= `1/2 xx 80 xx 11`

= 40 × 11

= 440 cm²

The area of trapezium PQRS is 440 cm².