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In the figure, ABCD is a rectangle, and the area of quadril

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In the figure, ABCD is a rectangle, and the area of quadril [#permalink] New post   17 Mar 2019, 12:41
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In the figure, ABCD is a rectangle, and the area of quadrilateral AFCE is equal to the area of \(∆ABC\). What is the value of x ?

(A) 5

(B) 6

(C) 7

(D) 12

(E) 15
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A
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MagooshStudentHelp
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Re: In the figure, ABCD is a rectangle, and the area of quadril [#permalink] New post   20 Mar 2019, 09:18
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Both quadrilateral AFCE and triangle ABC share triangle ACE.

This means that the parts the don't share, triangles AFC and EBC, must also have the same area.

If the base of triangle AFC is 3, then it's height is the width of the rectangle: 4.

Similarly, if the base of triangle EBC is 4, then it's height, EB, must be 3, so that it's area matches that of triangle AFC.

If EB = 3, then x = 5.
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Re: In the figure, ABCD is a rectangle, and the area of quadril [#permalink] New post   24 Oct 2021, 08:40
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Ar(AFC)+Ar(AEC)=Ar(ABC)

6+2x=16
x=5

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Re: In the figure, ABCD is a rectangle, and the area of quadril [#permalink] New post   04 Nov 2022, 07:37
MagooshStudentHelp wrote:
Both quadrilateral AFCE and triangle ABC share triangle ACE.

This means that the parts the don't share, triangles AFC and EBC, must also have the same area.

If the base of triangle AFC is 3, then it's height is the width of the rectangle: 4.

Similarly, if the base of triangle EBC is 4, then it's height, EB, must be 3, so that it's area matches that of triangle AFC.

If EB = 3, then x = 5.



Why Triangle EBC and ACF must have the same area ?
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Re: In the figure, ABCD is a rectangle, and the area of quadril [#permalink] New post   04 Nov 2022, 08:17
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The formula for the area of a triangle is 1/2 × base × height. Hence, the areas of triangles AEC, ACF, and ABC are

The area of ∆AEC = (1/2)(x)(BC) = (1/2)(x)(4) = 2x
The area of ∆ACF = (1/2)(3)(AD)
= (1/2)(3)(BC)

AD = BC, since opposite sides a rectangle are equal
= (3/2)BC = (3/2)(4) = 6

The area of ∆ABC = (1/2)(AB)(AD)
= (1/2)(CD)(BC)

AB = CD and AD = BC since opposite sides in a
rectangle are equal

= (1/2)(DF + FC)(BC)
= (1/2)(5 + 3)(4)
= (1/2)(8)(4)
= 16

Now, the area of the quadrilateral AECF equals (the area of ∆AEC) + (the area of ∆ACF) = 2x + 6, and the area of ∆ABC = 16. Since we are given that the area of the triangle ABC equals the area of the quadrilateral AFCE, 2x + 6 = 16. Solving for x in this equation yields x = 5. The answer is (A).
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