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In the figure, ABCD is a rectangle, and the area of quadril
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17 Mar 2019, 12:41
17 Mar 2019, 12:41
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83% (02:50) correct
16% (02:29) wrong based on 24 sessions
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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
Re: In the figure, ABCD is a rectangle, and the area of quadril
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20 Mar 2019, 09:18
20 Mar 2019, 09:18
Expert Reply
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.
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.
Re: In the figure, ABCD is a rectangle, and the area of quadril
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24 Oct 2021, 08:40
24 Oct 2021, 08:40
1
Ar(AFC)+Ar(AEC)=Ar(ABC)
6+2x=16
x=5
Posted from my mobile device
6+2x=16
x=5
Posted from my mobile device
