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Bunuel
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maybe a suggestion but: if experts solve this, could they wait till some people have posted their answers, or, at least encase their answer in spoiler tags?
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Bunuel

If square ABCD has area x^2 and line segment AE has length 4, and line segment AF has length 3, then what is the area of IGCH in terms of x ?

(A) 4x + 12
(B) 7x - 12
(C) x^2 - 24
(D) x^2 - 3 x + 12
(E) x^2 - 7 x + 12

Attachment:
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Attachment:
2018-01-18_1411.png
2018-01-18_1411.png [ 7.07 KiB | Viewed 3515 times ]

Area of IGCH = Area of ABCD - ( Area of AEFI + Area of FIBG + Area of EIDH )

Or, Area of IGCH = \(x^2 - ( 12 + 4x - 12 + 3x - 12 )\)

Or, Area of IGCH = \(x^2 - 7 x + 12\), answer will be (E)

aeon86
maybe a suggestion but: if experts solve this, could they wait till some people have posted their answers, or, at least encase their answer in spoiler tags?

Done , please solve this question now...
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Bunuel

If square ABCD has area x^2 and line segment AE has length 4, and line segment AF has length 3, then what is the area of IGCH in terms of x ?

(A) 4x + 12
(B) 7x - 12
(C) x^2 - 24
(D) x^2 - 3 x + 12
(E) x^2 - 7 x + 12

Attachment:
2018-01-18_1411.png

Since the area of the square ABCD is x^2, we observe that both AB and AD must have length x.

Since AB = x = EG and AF = 3 = EI, then IG = EG - EI = x - 3.

Since AD = x = FH and AE = 4 = FI, then IH = FH - FI = x - 4.

Thus, the area of IGCH is:

(x - 3)(x - 4) = x^2 - 7x + 12

Answer: E
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