ABCD is a trapezium with AD and BC as paralle | vedclass

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Question

(D) Let $EF$ be the perpendicular distance between the parallel sides $AD$ and $BC$,where $F$ lies on $AD$.The area of trapezium $ABCD = \frac{1}{2} \

Vedclass · Accepted Answer

$\frac{AD+BC}{AD}$

Vedclass · Answer

(D) Let $EF$ be the perpendicular distance between the parallel sides $AD$ and $BC$,where $F$ lies on $AD$.The area of trapezium $ABCD = \frac{1}{2} 	imes (AD + BC) 	imes EF$.The area of $\Delta AED = \frac{1}{2} 	imes AD 	imes EF$.Therefore,the required ratio is:$\frac{	ext{Area of } ABCD}{	ext{Area of } \Delta AED} = \frac{\frac{1}{2} 	imes (AD + BC) 	imes EF}{\frac{1}{2} 	imes AD 	imes EF} = \frac{AD + BC}{AD}$.

$ABCD$ is a trapezium with $AD$ and $BC$ as parallel sides. $E$ is a point on $BC$. The ratio of the area of $ABCD$ to that of $\Delta AED$ is

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