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Intern Joined: 19 Feb 2018
Posts: 5
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Kudos [?]: 0 , given: 3

In the square ABCD, PR is parallel to CD and the length of C [#permalink] 00:00

Question Stats: 62% (00:59) correct 37% (01:24) wrong based on 8 sessions
In the square ABCD, PR is parallel to CD and the length of CD is 10.

 Quantity A Quantity B Area of the shaded region 50

A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Attachments IMG_20180220_124727.jpg [ 773.85 KiB | Viewed 1682 times ]

Last edited by Carcass on 20 Feb 2018, 03:26, edited 1 time in total.
Edited by Carcass Manager  Joined: 15 Jan 2018
Posts: 147
GMAT 1: Q V
Followers: 3

Kudos [?]: 182  , given: 0

Re: In the square ABCD, PR is parallel to CD and the length of C [#permalink]
3
KUDOS
Line RP looks a little skewed. If this were an official ETS question, that little disconnect between the shaded region and line RP would mean there actually is a disconnect, making this a very different problem. I'm going to assume that the shaded area is meant to line up with line RP.

If so, then this can be a quite fast problem. Look at rectangle CRPD. Its area would be length DP x length PR. If you tilt your head to the left, you can imagine that PR is the base of the white triangle, whose area would be bh/2, which would be (length PR x length DP)/2. Thus, the white area is exactly half the rectangle. The same would be true of the white area on the right. So the two triangles take up exactly half the area of the square, the shaded area takes up the other half, and since the square's area is 100, the answer is C.

By the way, if the picture did intend for the triangles on the left to encroach to the right of line PR, the answer would be D. You can draw the triangles such that they still take exactly half the area, or you can skew them such that the shaded area takes up nearly the entire figure. Try it for yourself.
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Manager  Joined: 26 Jun 2017
Posts: 104
Followers: 0

Kudos [?]: 39 , given: 38

Re: In the square ABCD, PR is parallel to CD and the length of C [#permalink]
SherpaPrep wrote:
Line RP looks a little skewed. If this were an official ETS question, that little disconnect between the shaded region and line RP would mean there actually is a disconnect, making this a very different problem. I'm going to assume that the shaded area is meant to line up with line RP.

If so, then this can be a quite fast problem. Look at rectangle CRPD. Its area would be length DP x length PR. If you tilt your head to the left, you can imagine that PR is the base of the white triangle, whose area would be bh/2, which would be (length PR x length DP)/2. Thus, the white area is exactly half the rectangle. The same would be true of the white area on the right. So the two triangles take up exactly half the area of the square, the shaded area takes up the other half, and since the square's area is 100, the answer is C.

By the way, if the picture did intend for the triangles on the left to encroach to the right of line PR, the answer would be D. You can draw the triangles such that they still take exactly half the area, or you can skew them such that the shaded area takes up nearly the entire figure. Try it for yourself.

I really like the way you explain the problems
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What you think, you become.

Director Joined: 09 Nov 2018
Posts: 509
Followers: 0

Kudos [?]: 21 , given: 1

Re: In the square ABCD, PR is parallel to CD and the length of C [#permalink]
SherpaPrep wrote:
Line RP looks a little skewed. If this were an official ETS question, that little disconnect between the shaded region and line RP would mean there actually is a disconnect, making this a very different problem. I'm going to assume that the shaded area is meant to line up with line RP.

If so, then this can be a quite fast problem. Look at rectangle CRPD. Its area would be length DP x length PR. If you tilt your head to the left, you can imagine that PR is the base of the white triangle, whose area would be bh/2, which would be (length PR x length DP)/2. Thus, the white area is exactly half the rectangle. The same would be true of the white area on the right. So the two triangles take up exactly half the area of the square, the shaded area takes up the other half, and since the square's area is 100, the answer is C.

By the way, if the picture did intend for the triangles on the left to encroach to the right of line PR, the answer would be D. You can draw the triangles such that they still take exactly half the area, or you can skew them such that the shaded area takes up nearly the entire figure. Try it for yourself.

I think, RP is extraneous.
Intern Joined: 08 Dec 2018
Posts: 8
Followers: 0

Kudos [?]: 2 , given: 5

Re: In the square ABCD, PR is parallel to CD and the length of C [#permalink]
I want to put this way:
Area of the two triangles= (1/2*10*x) + ( 1/2*10* (10-x)) which equals 50. So, the answer is C .
Manager  Joined: 01 Nov 2018
Posts: 87
Followers: 0

Kudos [?]: 42 , given: 22

Re: In the square ABCD, PR is parallel to CD and the length of C [#permalink]
Expert's post
jelal123 wrote:
I want to put this way:
Area of the two triangles= (1/2*10*x) + ( 1/2*10* (10-x)) which equals 50. So, the answer is C .

I agree Re: In the square ABCD, PR is parallel to CD and the length of C   [#permalink] 21 Feb 2019, 23:13
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