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Points L, M, and, N have xy-coordinates (2,0), (8,12), and

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Points L, M, and, N have xy-coordinates (2,0), (8,12), and  [#permalink]

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New post 14 May 2012, 04:12
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Points L, M, and, N have xy-coordinates (2,0), (8,12), and (14,0), respectively. Points P, Q, and R have xy-coordinates (6,0), (8,4), and (10,0), respectively. What fraction of the area of the triangle LMN is the area of the triangle PQR?

A. \(\frac{1}{9}\)
B. \(\frac{1}{8}\)
C. \(\frac{1}{6}\)
D. \(\frac{1}{5}\)
E. \(\frac{1}{3}\)

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Re: Points L, M, and, N have xy-coordinates (2,0), (8,12), and  [#permalink]

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New post 14 May 2012, 04:35
Stiv wrote:
Points L, M, and, N have xy-coordinates (2,0), (8,12), and (14,0), respectively. Points P, Q, and R have xy-coordinates (6,0), (8,4), and (10,0), respectively. What fraction of the area of the triangle LMN is the area of the triangle PQR?

A. \(\frac{1}{9}\)
B. \(\frac{1}{8}\)
C. \(\frac{1}{6}\)
D. \(\frac{1}{5}\)
E. \(\frac{1}{3}\)


Look at the diagram below:
Attachment:
Triangles.png
Triangles.png [ 13.17 KiB | Viewed 3655 times ]


The area of triangle LMN (red) is 1/2*base*height=1/2*12*12=72;
The area of triangle PQR (blue) is 1/2*base*height=1/2*4*4=8;

The area of PQR is 8/72=1/9 of the area of triangle LMN.

Answer: A.

Else you can notice that triangles LMN and PQR are similar. Now, in two similar triangles, the ratio of their areas is the square of the ratio of their sides, therefore since the ratio of the sides is 4/12=1/3 then the ratio of the areas is (1/3)^2=1/9.

Hope it's clear.
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Re: Points L, M, and, N have xy-coordinates (2,0), (8,12), and  [#permalink]

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New post 25 Apr 2014, 07:27
Bunuel wrote:
Stiv wrote:
Points L, M, and, N have xy-coordinates (2,0), (8,12), and (14,0), respectively. Points P, Q, and R have xy-coordinates (6,0), (8,4), and (10,0), respectively. What fraction of the area of the triangle LMN is the area of the triangle PQR?

A. \(\frac{1}{9}\)
B. \(\frac{1}{8}\)
C. \(\frac{1}{6}\)
D. \(\frac{1}{5}\)
E. \(\frac{1}{3}\)


Look at the diagram below:
Attachment:
Triangles.png


The area of triangle LMN (red) is 1/2*base*height=1/2*12*12=72;
The area of triangle PQR (blue) is 1/2*base*height=1/2*4*4=8;

The area of PQR is 8/72=1/9 of the area of triangle LMN.

Answer: A.

Else you can notice that triangles LMN and PQR are similar. Now, in two similar triangles, the ratio of their areas is the square of the ratio of their sides, therefore since the ratio of the sides is 4/12=1/3 then the ratio of the areas is (1/3)^2=1/9.

Hope it's clear.



Hi,
I think I won't try immediately to draw diagram in real exam. So it would be very helpful if you give formula to find area from the points if there is any. Please help me to solve the problem without drawing anything.
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Re: Points L, M, and, N have xy-coordinates (2,0), (8,12), and  [#permalink]

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New post 25 Apr 2014, 18:42
1
If you didn't want to draw, you could notice that both triangles have bases that run along the x-axis. The first triangle has a base that runs from (2,0) to (14,0) and a vertex at (8,12), so that's a base of 12 and a height of 12. The second triangle has a base that runs from (6,0) to (10,0) and a vertex at (8,4), so that's a base of 4 and a height of 4. The second triangle is 1/3 as big in both directions, so it will have 1/9 the area of the first one. (1/3 * 1/3)

Having said all that, most people would probably do best to draw a picture. It doesn't have to be pretty and it should only take a few seconds to make. If it ensures that you visualize the problem correctly, then it's worth it. The step you can skip is actually calculating the areas. You only need to know PQR's area relative to LMN; you don't have to know the actual values.
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Re: Points L, M, and, N have xy-coordinates (2,0), (8,12), and  [#permalink]

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New post 25 Apr 2014, 19:11
DmitryFarber wrote:
If you didn't want to draw, you could notice that both triangles have bases that run along the x-axis. The first triangle has a base that runs from (2,0) to (14,0) and a vertex at (8,12), so that's a base of 12 and a height of 12. The second triangle has a base that runs from (6,0) to (10,0) and a vertex at (8,4), so that's a base of 4 and a height of 4. The second triangle is 1/3 as big in both directions, so it will have 1/9 the area of the first one. (1/3 * 1/3)

Having said all that, most people would probably do best to draw a picture. It doesn't have to be pretty and it should only take a few seconds to make. If it ensures that you visualize the problem correctly, then it's worth it. The step you can skip is actually calculating the areas. You only need to know PQR's area relative to LMN; you don't have to know the actual values.



Thank you very much. The explanation is helpful.
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Re: Points L, M, and, N have xy-coordinates (2,0), (8,12), and  [#permalink]

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New post 27 Apr 2014, 11:52
This is a problem from GMATPrep that does not include the first two sentences posted by the original poster, and instead has an image in the coordinate plane. Please see the attached image and use that to solve this problem as opposed to modifying the original problem statement.

Cheers,
Dabral
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Test1-Q05-PS01.png
Test1-Q05-PS01.png [ 121.54 KiB | Viewed 2308 times ]

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Re: Points L, M, and, N have xy-coordinates (2,0), (8,12), and  [#permalink]

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New post 28 Apr 2014, 02:42
Raihanuddin wrote:
Bunuel wrote:
Stiv wrote:
Points L, M, and, N have xy-coordinates (2,0), (8,12), and (14,0), respectively. Points P, Q, and R have xy-coordinates (6,0), (8,4), and (10,0), respectively. What fraction of the area of the triangle LMN is the area of the triangle PQR?

A. \(\frac{1}{9}\)
B. \(\frac{1}{8}\)
C. \(\frac{1}{6}\)
D. \(\frac{1}{5}\)
E. \(\frac{1}{3}\)


Look at the diagram below:
Attachment:
Triangles.png


The area of triangle LMN (red) is 1/2*base*height=1/2*12*12=72;
The area of triangle PQR (blue) is 1/2*base*height=1/2*4*4=8;

The area of PQR is 8/72=1/9 of the area of triangle LMN.

Answer: A.

Else you can notice that triangles LMN and PQR are similar. Now, in two similar triangles, the ratio of their areas is the square of the ratio of their sides, therefore since the ratio of the sides is 4/12=1/3 then the ratio of the areas is (1/3)^2=1/9.

Hope it's clear.



Hi,
I think I won't try immediately to draw diagram in real exam. So it would be very helpful if you give formula to find area from the points if there is any. Please help me to solve the problem without drawing anything.


I believe formulae is there but it will the solution time-consuming. Definitely not advisable for a strictly timed exam like GMAT
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Re: Points L, M, and, N have xy-coordinates (2,0), (8,12), and  [#permalink]

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Re: Points L, M, and, N have xy-coordinates (2,0), (8,12), and &nbs [#permalink] 29 Nov 2018, 22:22
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