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,


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


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