Find the area of the trapezium PQRS given in | vedclass

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(A) To find the area of the trapezium $PQRS$,draw a line $ST$ perpendicular to $PQ$ at $T$. Since $PQ \parallel SR$ and $QR \perp PQ$,$QR \perp SR$ as

Vedclass · Accepted Answer

$114$

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(A) To find the area of the trapezium $PQRS$,draw a line $ST$ perpendicular to $PQ$ at $T$. Since $PQ \parallel SR$ and $QR \perp PQ$,$QR \perp SR$ as well. Thus,$TQRS$ is a rectangle. Given $PQ = 12 \, m$ and $SR = 7 \, m$,we have $TQ = SR = 7 \, m$. Therefore,$PT = PQ - TQ = 12 \, m - 7 \, m = 5 \, m$. In the right-angled triangle $	riangle PTS$,by Pythagoras theorem: $TS^2 = PS^2 - PT^2 = 13^2 - 5^2 = 169 - 25 = 144$. So,$TS = \sqrt{144} = 12 \, m$. Since $TQRS$ is a rectangle,$QR = TS = 12 \, m$. The area of the trapezium is given by the formula: $	ext{Area} = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$ $	ext{Area} = \frac{1}{2} 	imes (PQ + SR) 	imes QR$ $	ext{Area} = \frac{1}{2} 	imes (12 + 7) 	imes 12 = 19 	imes 6 = 114 \, m^2$.

Find the area of the trapezium $PQRS$ given in the figure.

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