A field is in the shape of a trapezium having | vedclass

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Question

(A) Let the trapezium be $ABCD$ with parallel sides $AB = 90 \, m$ and $CD = 30 \, m$. The side $AC$ is perpendicular to both $AB$ and $CD$.Draw $DM \

Vedclass · Accepted Answer

$19200$

Vedclass · Answer

(A) Let the trapezium be $ABCD$ with parallel sides $AB = 90 \, m$ and $CD = 30 \, m$. The side $AC$ is perpendicular to both $AB$ and $CD$.Draw $DM \perp AB$. Since $ACDM$ forms a rectangle,$AM = CD = 30 \, m$ and $DM = AC$.Now,$MB = AB - AM = 90 \, m - 30 \, m = 60 \, m$.In the right-angled triangle $	riangle DMB$,by Pythagoras theorem:$DM^2 = DB^2 - MB^2 = (100)^2 - (60)^2$$DM^2 = 10000 - 3600 = 6400$$DM = \sqrt{6400} = 80 \, m$.Thus,the height of the trapezium is $80 \, m$.The area of the trapezium $ABCD = \frac{1}{2} 	imes (	ext{sum of parallel sides}) 	imes 	ext{height}$$= \frac{1}{2} 	imes (90 + 30) 	imes 80 = \frac{1}{2} 	imes 120 	imes 80 = 4800 \, m^2$.Total cost of ploughing = $	ext{Area} 	imes 	ext{Rate} = 4800 	imes 4 = \operatorname{Rs} 19,200$.

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