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Picture Frame Inscribed Problem

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Picture Frame Inscribed Problem

by sanchu » Tue Feb 25, 2014 4:32 pm
Manhattan GMAT number 10

ABCD is a a square picture frame. EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH is equal to the area of the picture frame. (The area of ABCD minus EFGH). If AB=6. What is the length of EF?

I know the total area is 36, but how do I deduce the area of EFGH?
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by GMATGuruNY » Tue Feb 25, 2014 6:44 pm
sanchu wrote:Manhattan GMAT number 10

ABCD is a a square picture frame. EFGH is a square inscribed within ABCD as a space for a picture. The area of EFGH is equal to the area of the picture frame. (The area of ABCD minus EFGH). If AB=6. What is the length of EF?

I know the total area is 36, but how do I deduce the area of EFGH?
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Square ABCD:
Since AB=6, area = 6² = 36.

Square EFGH:
Since the area of EFGH is equal to the area of the surrounding frame, EFGH constitutes 1/2 the total area of ABCD, while the surrounding frame constitutes the remaining 1/2.
Thus, the area of EFGH = (1/2)(36) = 18.
Since s² = area, we get:
s² = 18
s = √18 = √9 * √2 = 3√2.
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by [email protected] » Tue Feb 25, 2014 8:44 pm
Hi sanchu,

In this prompt, we're told that a square is inscribed in another square. With the rest of the info in the prompt, we know the following:

1) (area of small square) = (area of "frame")
2) (area of small square) + (area of "frame") = (area of big square) = 36

With these two formulas, we know that the small square and the frame are EQUAL and that they add up to 36. So, they must both = 18

Area of a square = S^2 so....

S^2 = 18

S = 3(root2)

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