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In the diagram to the right, triangle PQR has a right angle [#permalink]

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31 Oct 2007, 09:07

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In the diagram to the right, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?

In the diagram to the right, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?

3/2

7/4

15/8

16/9

2

i get 16/9 but after spending too much time trying to figure it out.
i'll try my best to explain as best as i can (see my diagram below)

stem only tells you 1) these three triangles are all right triangles (so you can easily apply pythagorean theorem), 2) the perimeter of the largest triangle, 3) the length of the bisector, and 4) PQ > QR. From this you have to think about the possible lengths to find the sides.

I thought about the common right triangle sides: 3-4-5, 6-8-10, etc and saw that 3-4-5 = 12 and 60 is a multiple of 12 (5x). So the possible sides are [3-4-5]*5 = 15-20-25. But first have to test it out. We know that PQ > QR and the largest side is the hypotenuse. So PQ = 20, QR = 15, and PR = 25. From there I used side QS = 12 to figure out the splits between PS and RS. (Lucky thing it worked out!!)

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]

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15 Apr 2012, 21:52

jimjohn wrote:

In the diagram to the right, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS?

3/2

7/4

15/8

16/9

2

can somebody give OA for this? _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Re: In the diagram to the right, triangle PQR has a right angle [#permalink]

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16 Apr 2012, 00:45

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harshavmrg wrote:

can somebody give OA for this?

In the diagram, triangle PQR has a right angle at Q and a perimeter of 60. Line segment QS is perpendicular to PR and has a length of 12. PQ > QR. What is the ratio of the area of triangle PQS to the area of triangle RQS? A. 3/2 B. 7/4 C. 15/8 D. 16/9 E. 2

Attachment:

Triangle PQR.GIF [ 2.52 KiB | Viewed 3928 times ]

Let \(PQ=x\), \(QR=y\) and \(PR=z\).

Given: \(x+y+z=60\) (i); Equate the areas: \(\frac{1}{2}*xy=\frac{1}{2}*QS*z\) (area of PQR can be calculated by 1/2*leg*leg and 1/2* perpendicular to hypotenuse*hypotenuse) --> \(xy=12z\) (ii); Aslo \(x^2+y^2=z^2\) (iii);

So, we have: (i) \(x+y+z=60\); (ii) \(xy=12z\); (iii) \(x^2+y^2=z^2\).

From (iii) \((x+y)^2-2xy=z^2\) --> as from (i) \(x+y=60-z\) and from (ii) \(xy=12z\) then (\(60-z)^2-2*12z=z^2\) --> \(3600-120z+z^2-24z=z^2\) --> \(3600=144z\) --> \(z=25\);

From (i) \(x+y=35\) and from (ii) \(xy=300\) --> solving for \(x\) and \(y\) --> \(x=20\) and \(y=15\) (as given that \(x>y\)).

Next, perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. So, PQR and SQR are similar. In two similar triangles, the ratio of their areas is the square of the ratio of their sides: \(\frac{AREA}{area}=\frac{S^2}{s^2}\).

So, \(\frac{x^2}{y^2}=\frac{AREA}{area}\) --> \(\frac{AREA}{area}=\frac{400}{225}=\frac{16}{9}\)

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