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Re: In the figure above, ABCD is a parallelogram and E [#permalink]
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The perspicuity of your replies always surprises me!! GreenlightTestPrep
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In the figure above, ABCD is a parallelogram and E [#permalink]
2
Given ABCD is a parallelogram

Let AD=BC = b {base} ; height of parallelogram ABCD be "h"

So,area of parallelogram ABCD = base * height= \(bh\)

The figure BCDE is trapezium, whose area is calculated using formula \(\frac{1}{2}\)\(h(a+b)\)
=\(\frac{1}{2}\)\(h(b+\)\(\frac{b}{2}\))
=\(\frac{3}{4}\)\(bh\)

Area of triangle ABE = Area of parallelogram ABCD - Area of trapezium BCDE
=\(\frac{1}{4}\)\(bh\)

We were asked to find
\(\frac{Area of Triangle ABE}{Area of Trapezium BCDE}\)

plug-in above deduced values, we get
\(\frac{Area of Triangle ABE}{Area of Trapezium BCDE}\) = \(\frac{1}{3}\)

Therefore the answer is Option B
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Re: In the figure above, ABCD is a parallelogram and E [#permalink]
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Re: In the figure above, ABCD is a parallelogram and E [#permalink]
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