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