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

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#### Solution

We have, trapezium PQRD, 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 Pythagoras theorem]

⇒ (13)^{2} = (5)^{2} + (TR)^{2} .......[By Pythagoras theorem]

⇒ (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 m^{2}

Now, Area of rectangle PQRT = PQ x RQ = 12 × 7 ......[∵ Area of a rectangle = length × breadth]

= 84 m^{2} .....[∵ PQ = TR = 12 m]

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

= 30 + 84

= 114 m^{2}

Hence, the area of trapezium is 114 m^{2}.

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