Triangle question from online exam

[C129]

Can someone please explain how to work the below question? The explanation from answer was not very clear to me. Thank you.

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

[RonPurewal]

could you please post the diagram from this problem? the problem statement isn't much help without it.

you can use one of many free image hosting sites, such as supload.com or postimage.org, to create a link to the image for your post.

thanks.

[C129]

http://www.postimage.org/image.php?v=aVzed6i

[thesamet]

It's impossible to solve the question in an algebraic way in 2 minutes (unless there's an insight that I am missing). Since the small triangles share the same height, we are asked for the ratio of their bases (PS/SR). The nice ratio between the height (12) and the perimeter (60) made me *guess* that all sides of the triangles are integers. Assuming the edges ratio are 3:4:5, and the perimeter is 60, it must be a 15:20:25 triangle. For Pythagorean equation leads to PS=16 and SR=9. Since 16:9 matches an answer choice, the "guessamption" was right.

Here is an algebraic solution which I find easier. For a right triangle with sides a and b and hypotenuse c, there is a nice formula connecting its perimeter p with its area S:

4S = p^2 - 2pc

It has a simple proof:
p^2 - 2pc = a^2+b^2+c^2 + 2ac + 2bc + 2ab - 2(a+b+c)c = a^2+b^2+c^2+2ab-2c^2 = 2ab

The last equation holds since a^2+b^2=c^2. Now since 2ab is 4S we are done. Now, for the original question:

Let a = |PS| and b=|SR|. We need to find a/b. Since the two triangles are similar, we have 12/a=b/12, so we conclude that ab=144.

The area of the external triangle is 12(a+b)/2. Let's use the formula above:

24(a+b) = 3600 - 120(a+b)
144(a+b) = 3600
a+b=25

so a+b=25 and a*b=144 implies that a=16 and b=9 (a=9 b=16 is impossible since we are told which triangle is the largest).

[esledge]

Personally, I would "solve" as thesamet did, using the suspicion of a 3-4-5 triangle as a guide. It's not as risky or outlandish as it might seem: the GMAT often employs 3-4-5 triangles, the numbers are quickly checked by formulas, and finding the answer among the PS choices provides a final check. (Side note, just to cover everything: I'm just thinking that such an approach could backfire on DS, as you could easily slip into using that assumption as fact...)