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

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Joined: 19 Aug 2007
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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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63% (04:23) correct 38% (03:35) wrong based on 152 sessions

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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?

A.. 3/2
B. 7/4
C. 15/8
D. 16/9
E. 2
Attachment:

triangle.jpg [ 6.39 KiB | Viewed 4886 times ]

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-diagram-triangle-pqr-has-a-right-angle-at-q-and-a-127093.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 16 Apr 2012, 00:46, edited 1 time in total.
Edited the question and added the OA. Topic is locked.
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31 Oct 2007, 09:47
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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

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!!)

PS^2 + QS^2 = PQ^2
PS^2 = PQ^2 - QS^2
400 - 144 = 256
sqrt256 = 16

QS^2 + RS^2 = QR^2
RS^2 = QR^2 - QS^2
225 - 144 = 81
sqrt 81 = 9

so the ratio of the area of triangle PQS to the area of triangle RQS is
PQS = 1/2bh
1/2 * 12 * 16 = 96

RQS = 1/2bh
1/2 * 12 * 9 = 54

96:54 = 16:9
Attachments

geometry.JPG [ 9.88 KiB | Viewed 4838 times ]

Last edited by beckee529 on 31 Oct 2007, 09:54, edited 2 times in total.
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31 Oct 2007, 09:53
Yeah, and actually, your calculations were still correct. I made an error in my calculations, the answer is not 2. It is 16/9 as you stated.
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31 Oct 2007, 09:55
emoryhopeful wrote:
Yeah, and actually, your calculations were still correct. I made an error in my calculations, the answer is not 2. It is 16/9 as you stated.

yeah i had it right on paper but wrote it incorrectly on the computer screen but have since edited to correct it
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Re: In the diagram to the right, triangle PQR has a right angle [#permalink]

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15 Apr 2012, 18:27

we've given that the hypotenuse of triangle PQR is 25.
Applying Pythagorean triplet 15-20-25,

PQ and QR are respectively 20 and 15.

So area = 1/2 * base QR * height PQ
= 1/2 * 15 * 20
= 150
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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?
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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

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Posts: 39589
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 4607 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, $$\frac{x^2}{y^2}=\frac{AREA}{area}$$ --> $$\frac{AREA}{area}=\frac{400}{225}=\frac{16}{9}$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-diagram-triangle-pqr-has-a-right-angle-at-q-and-a-127093.html
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Re: In the diagram to the right, triangle PQR has a right angle   [#permalink] 16 Apr 2012, 00:45
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