Q. 10

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

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,

RS^{2} = ST^{2} + RT^{2}

⇒ 13^{2} = 12^{2} + RT^{2}

⇒ 169 = 144 + RT^{2}

⇒ RT^{2} = 169-144

⇒ RT^{2} = 25

⇒ RT = 5m

Area (PQRS) =

Area(PQRS) =

Area(PQRS) = 9.5 × 5 = 47.5 m^{2}

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RS Aggarwal & V Aggarwal - Mathematics