Find the area of the trapezium PQRS with height PQ given in Fig. 12.3

Question

Philoid · Accepted Answer

We draw perpendicular from R on PS, which is also parallel to PQ.

PQRT is a rectangle.

ST = PS – TP

ST = 12 – 7 = 5m (TP = RQ = 7m)

In Δ RTS,

RS² = ST² + RT²

⇒ 13² = 12² + RT²

⇒ 169 = 144 + RT²

⇒ RT² = 169-144

⇒ RT² = 25

⇒ RT = 5m

Area (PQRS) =

Area(PQRS) =

Area(PQRS) = 9.5 × 5 = 47.5 m²