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Bunuel
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Let AC and DE meet at O. So, triangle AOE and triangle COD are similar.

=> Sides and corresponding heights will be in the same ratio.

Say, COD height h1, AOE height h2. => AE/CD = h2/h1 => h2/h1 = 1/2
But, h1+h2 = AD = CD => h2 = (2/3) * CD

Area of triangle COD = 0.5 * h2 * CD = 0.5 * (2/3)*CD * CD = (1/3) * CD^2 = 1/3 * Area of square

Ratio= 1/3
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Kindly see the attachment.
D
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1.PNG [ 60.77 KiB | Viewed 5858 times ]

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Bunuel

What fraction of the square ABCD above, where AE = EB, is shaded?

A. 1/6
B. 1/5
C. 1/4
D. 1/3
E. 1/2

Ans is D .
Please get the attachment for the solution .
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File comment: Solution diagram
17_june.PNG
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vtexcelgmat
I want to share a simple solution WHICH DOES NOT INVOLVE MAKING ANY SKETCH/DIAGRAM, but a tiny little imagination.

The shaded area is ‘less than triangle ADC’ therefore less than ½ of area of square.
Imagine diagonal line joining Point D and B. Let us say the point of intersection of two diagonals is O.
Now, the shaded area is ‘more than triangle DOC’ therefore more than ¼ of area of square.

Out of 5 options, how many options have value between ½ and ¼ ? Just 1/3.

This is a brilliant answer. Kudos.
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Why can't you subtract the area of ADE from ACD to get the area of shaded region?
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Hello from the GMAT Club BumpBot!

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