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505-555 (Easy)|   Geometry|      
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Bunuel
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If square ABCD has area x^2 and line segment AE has length 4, and line 18 Jan 2018, 03:13
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90% (01:45) correct 10% (01:49) wrong based on 49 sessions

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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

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E
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If square ABCD has area x^2 and line segment AE has length 4, and line 18 Jan 2018, 04:53
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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) 4 x + 12
(B) 7 x - 12
(C) x^2 - 24
(D) x^2 - 3 x + 12
(E) x^2 - 7 x + 12

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2018-01-18_1411.png
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Since the calculation looks straightforward, we'll just do it.
This is a Precise approach.

As the area of square ABCD is x^2 its side has length x.
Then IH = x - AE = x - 4 and IG = x - AF = x - 3
So our area is (x-4)(x-3) which is answer (E).

Bunuel I assume that x2 in the question should be x^2?
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If square ABCD has area x^2 and line segment AE has length 4, and line 18 Jan 2018, 06:01
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DavidTutorexamPAL
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) 4 x + 12
(B) 7 x - 12
(C) x^2 - 24
(D) x^2 - 3 x + 12
(E) x^2 - 7 x + 12

Show Spoiler
Attachment:
2018-01-18_1411.png

Since the calculation looks straightforward, we'll just do it.
This is a Precise approach.

As the area of square ABCD is x^2 its side has length x.
Then IH = x - AE = x - 4 and IG = x - AF = x - 3
So our area is (x-4)(x-3) which is answer (E).

Bunuel I assume that x2 in the question should be x^2?
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Yes. Edited. Thank you for noticing.
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If square ABCD has area x^2 and line segment AE has length 4, and line 18 Jan 2018, 06:05
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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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If square ABCD has area x^2 and line segment AE has length 4, and line 18 Jan 2018, 07:35
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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

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Attachment:
The attachment 2018-01-18_1411.png is no longer available
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2018-01-18_1411.png
2018-01-18_1411.png [ 7.07 KiB | Viewed 3501 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?
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Done , please solve this question now...
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If square ABCD has area x^2 and line segment AE has length 4, and line 20 Jan 2018, 07:32
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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

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Attachment:
2018-01-18_1411.png
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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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