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Re: In the figure shown ABCD is a rectangle. The shaded region represents [#permalink]
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Susarla96 wrote:
chetan2u wrote:
KeyurJoshi wrote:

In the figure shown ABCD is a rectangle. The shaded region represents what percentage of rectangle ABCD?

(1) AD : BD = 5 : 12
(2) EF : BD = 1 : 3



Pretty straightforward..
Now, \(\triangle ABD\) and \(\triangle EFA\) have SAME height, say h, so the area depends only on the base, that is EF and BD.
\(\frac{A(\triangle EFA)}{A(\triangle ABD)}=\frac{\frac{1}{2}*h*EF}{\frac{1}{2}*h*BD}=\frac{EF}{BD}\)
Area of rectangle is twice that of \(\triangle ABD\).
So our answer = \(\frac{A(\triangle EFA)}{2*A(\triangle ABD)}=\frac{\frac{1}{2}*h*EF}{2*\frac{1}{2}*h*BD}=\frac{EF}{2*BD}\)

(1) AD : BD = 5 : 12
We cannot say the exact lengths of any of the lines or location of E and F.
We can just say: Ratio of the sides of triangles so formed by the diagonal is 5:12:13

(2) EF : BD = 1 : 3
Gives us \(\frac{EF}{BD}=\frac{1}{3}\).
\(\frac{EF}{2*BD}=\frac{1}{2*3}=\frac{1}{6}\)
Sufficient


B


Hey "chetan2u" thank you for the explanation.

I understood every other part of the answer except for one thing: How come the base of the triangle ABD is BD, shouldn't be AB?

Please let me know. TIA :)



HI,

Base of the triangle can be any out of AB, BD or DA. In this case, taking it BD allows us to conclude that shaded region is 1/3 of triangle ABD and hence 1/6 of the rectangle.
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Re: In the figure shown ABCD is a rectangle. The shaded region represents [#permalink]
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