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In the figure, ABCD is a rectangle and AF is parallel to BE
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24 Mar 2019, 00:36
24 Mar 2019, 00:36
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90% (01:36) correct
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#GREpracticequestion In the figure, ABCD.jpg [ 20.56 KiB | Viewed 3589 times ]
In the figure, ABCD is a rectangle and AF is parallel to BE. If x = 5, and y = 10, then what is the area of \(∆AFD\) ?
(A) 2.5
(B) 5
(C) 12.5
(D) 50
(E) 50 + 5y
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Re: In the figure, ABCD is a rectangle and AF is parallel to BE
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07 Jun 2020, 04:08
07 Jun 2020, 04:08
Concept tested is Congruent Triangles
Two Triangles are Congruent (Same dimensions) if one side and two included angles are equal
Triangle ADF and Triangle BCE are congruent because AD = BC, Angle ADB = Angle BCE, Angle DAF = Angle CBE (As AF is parallel to BE)
So, DF = CE
Area of Triangle ADF = \(\frac{1}{2} * AD*DF\) = \(\frac{1}{2} *x*x\) = \(\frac{1}{2}5*5\)
= \(\frac{25}{2}\) = 12.5
So, answer will be C
Hope it helps!
Two Triangles are Congruent (Same dimensions) if one side and two included angles are equal
Triangle ADF and Triangle BCE are congruent because AD = BC, Angle ADB = Angle BCE, Angle DAF = Angle CBE (As AF is parallel to BE)
So, DF = CE
Area of Triangle ADF = \(\frac{1}{2} * AD*DF\) = \(\frac{1}{2} *x*x\) = \(\frac{1}{2}5*5\)
= \(\frac{25}{2}\) = 12.5
So, answer will be C
Hope it helps!
gmatclubot
Re: In the figure, ABCD is a rectangle and AF is parallel to BE [#permalink]
07 Jun 2020, 04:08
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