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In the square ABCD, PR is parallel to CD and the length of C
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Updated on: 20 Feb 2018, 03:26
Updated on: 20 Feb 2018, 03:26
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72% (01:30) correct
27% (01:24) wrong based on 11 sessions
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In the square ABCD, PR is parallel to CD and the length of CD is 10.
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.
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.
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Originally posted by Haripriya517 on 19 Feb 2018, 23:23.
Last edited by Carcass on 20 Feb 2018, 03:26, edited 1 time in total.
Last edited by Carcass on 20 Feb 2018, 03:26, edited 1 time in total.
Edited by Carcass
Sherpa Prep Representative
Joined: 15 Jan 2018
Posts: 147
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Re: In the square ABCD, PR is parallel to CD and the length of C
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20 Feb 2018, 12:25
20 Feb 2018, 12:25
3
Expert Reply
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.
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.
Re: In the square ABCD, PR is parallel to CD and the length of C
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04 Mar 2018, 05:03
04 Mar 2018, 05:03
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.
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
