Question

Find the area of the trapezium PQRS with height PQ given in the following figure.

Answer 1

We have, trapezium PQRS, in which draw a line RT perpendicular to PS.

Where, side, ST = PS – TP = 12 – 7 = 5 m.  ...[∵ TP = PQ = 7 m]

In right angled ΔSTR,

(SR)² = (ST)² + (TR)²   ...[By using Pythagoras theorem]

⇒ (13)² = (5)² + (TR)²  

⇒ (TR)² = 169 – 25

⇒ (TR)² = 144

∴ TR = 12 m   ...[Taking positive square root because length is always positive]

Now, area of ΔSTR = `1/2 xx TR xx TS`  ...[∵ Area of triangle = `1/2` (base × height)]

= `1/2 xx 12 xx 5`

= 30 m²

Now, area of rectangle PQRT = PQ × RQ  ...[∵ Area of a rectangle = length × breadth]

= 12 × 7 

= 84 m²   ...[∵ PQ = TR = 12 m]

∴ Area of trapezium = Area of ΔSTR + Area of rectangle PQRT

= 30 + 84

= 114 m²

Hence, the area of trapezium is 114 m².