Find the area of the trapezium PQRS with heig | vedclass

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(A) Draw $RT \perp SP$. From the figure,it is clear that $RT = PQ = 7 \ m$ and $PT = QR = 7 \ m$.Since $SP = 12 \ m$,we have $ST = SP - PT = 12 \ m -

Vedclass · Accepted Answer

$114$

Vedclass · Answer

(A) Draw $RT \perp SP$. From the figure,it is clear that $RT = PQ = 7 \ m$ and $PT = QR = 7 \ m$.Since $SP = 12 \ m$,we have $ST = SP - PT = 12 \ m - 7 \ m = 5 \ m$.Now,in the right-angled triangle $STR$,by Pythagoras theorem:$SR^2 = ST^2 + RT^2$$13^2 = 5^2 + RT^2$$169 = 25 + RT^2$$RT^2 = 169 - 25 = 144$$RT = 12 \ m$.Thus,the height of the trapezium is $12 \ m$.The area of trapezium $PQRS = \frac{1}{2} 	imes (SP + QR) 	imes PQ$$= \frac{1}{2} 	imes (12 \ m + 7 \ m) 	imes 12 \ m$$= \frac{1}{2} 	imes 19 \ m 	imes 12 \ m = 19 	imes 6 \ m^2 = 114 \ m^2$.

Find the area of the trapezium $PQRS$ with height $PQ$ given in the figure (in $m^2$).

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