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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Picture Frame Inscribed Problem
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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?
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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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)
GMAT assassins aren't born, they're made,
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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)
GMAT assassins aren't born, they're made,
Rich
