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In the figure, ABCD is a rectangle. Points P, Q, R, S, and T
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19 Mar 2019, 05:47
19 Mar 2019, 05:47
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In the figure, ABCD is a rectangle. Points P, Q, R, S, and T cut side AB of the rectangle such that AP = 3, PQ = QR = RS = ST = 1. E is a point on AD such that AE = 3. Which one of the following line segments is parallel to the diagonal BD of the rectangle?
(A) EP
(B) EQ
(C) ER
(D) ES
(E) ET
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Re: In the figure, ABCD is a rectangle. Points P, Q, R, S, and T
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07 Jun 2020, 02:25
07 Jun 2020, 02:25
1
This question is based on the concept of Similar triangles.
If Two triangles are similar then their corresponding angles are equal.(And if we are able to prove that the angles are equal then we can prove that the sides are parallel)
We need to find a triangle which is similar to triangle ADB, which will be triangle AEX (We need to find which vertex is X -> P/Q/R/S/T)
We are going to use the property of similar triangle to find the similar triangle
If two triangles are similar then their sides will be in the same ratio.
\(\frac{AE}{AD}\) = \(\frac{AX}{AB}\)
=> AX = \(\frac{AB*AE}{AD}\) = \(\frac{15*3}{9}\) = 5
=> AX = 5 => X is nothing but R
So, AER ~ ADB
=> Angle ARE = Angle ABD
=> ER || DB
So, answer will be C
Hope it helps!
If Two triangles are similar then their corresponding angles are equal.(And if we are able to prove that the angles are equal then we can prove that the sides are parallel)
We need to find a triangle which is similar to triangle ADB, which will be triangle AEX (We need to find which vertex is X -> P/Q/R/S/T)
We are going to use the property of similar triangle to find the similar triangle
If two triangles are similar then their sides will be in the same ratio.
\(\frac{AE}{AD}\) = \(\frac{AX}{AB}\)
=> AX = \(\frac{AB*AE}{AD}\) = \(\frac{15*3}{9}\) = 5
=> AX = 5 => X is nothing but R
So, AER ~ ADB
=> Angle ARE = Angle ABD
=> ER || DB
So, answer will be C
Hope it helps!
gmatclubot
Re: In the figure, ABCD is a rectangle. Points P, Q, R, S, and T [#permalink]
07 Jun 2020, 02:25
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