In the diagram to the right, triangle PQR

[garylyon]

In the diagram to the right, triangle PQR has a right angle at Q and line segment QS is perpendicular to PR. If line segment PS has a length of 16 and line segment SR has a length of 9, what is the area of triangle PQR?

72
96
108
150
200

For this problem, I used the 3-4-5 relationship to solve it quickly, which is not listed in the answer. I have this itching suspicioun that I got lucky - that this wouldn't necessarily hold true just because line PR is a multiple of 5 and PQR is a right triangle. I know my suspicioun would be well founded if we were talking about bases. For example, a right triangle with a base of 8 could be 3-4-5 or 8-15-17, I suppose. But we know its the hypotenuse, since it's opposite the right angle. What do you guys think?

[tim]

that's a good hunch to pursue if you don't have a better plan for solving the problem. you'd be surprised how often triangles on the GMAT fit into these narrow categories. of course, your best bet is to verify this information by solving for the altitude using similar triangles..

[asth678]

I got this sum wrong but I had the same hunch since PR which is the hypotenuse of the triangle is 25 (16+9=25) its a 7-24-25 triangle.Should the area be 1/2*24*7.This was not one of teh answer choices.Why is this wrong

What should my though process be for such questions-Should I always be looking for triangles which are similar?

Also the soln says-
Since QSP, PQR, and QSR are all right angles, we know that:

w + x = 90
x + y = 90
Therefore, w = y

Just double checking you get w=y by subt eq 1 from 2 right?you get w-y=0 therefore w=y.Same logic is used below?

x + y = 90
y + z = 90
Therefore, x = z

[tim]

You are correct about how you get w = y and x = z. As for your hunch, you can't seriously be asking why the answer you got based on a hunch was wrong! :) It was a hunch - sometimes those work out and sometimes they don't. The real way to do this problem is to notice three similar triangles and set up proportions.

[BradK592]

If angle PQR is a right angle, and the line segment QS is perpendicular to line PR, can we infer that angles PQS and SQR are 45 degrees each (i.e. cut in half by line QS)? Further, if angle PQS is 45 degrees, than angle QPR, must be 45 degrees (since we already know that angle QPS is 90 degrees). Then we have a 45-45-90 triangle and since line PS is 16, then line QS must be 16 (using the rules for 45-45-90 triangles--> x:x:x*(sq root 2)). Using this logic, results in the wrong answer (area=1/2(16)(25)=200), but I am not clear on why that doesn't work.

[RonPurewal]

Please post the diagram that accompanies the problem.

If you don't know how to upload an image directly, you can post the image to an image-hosting website, and post the link here.

Thank you.

[Shan_in_SS]

[RonPurewal]

Shan, thanks for posting the picture.

[RonPurewal]

If angle PQR is a right angle, and the line segment QS is perpendicular to line PR, can we infer that angles PQS and SQR are 45 degrees each (i.e. cut in half by line QS)?

No. That's true only if you are already starting from a 45º-45º-90º triangle, which is pretty clearly not the case in this problem.

In general, you'll get smaller triangles with exactly the same three angle measures as the one you started with. (To show this, note that each smaller triangle will share one of the acute angles with the original right triangle"”and that each smaller triangle, like the original, contains a 90º angle. If two of the angles are the same, then all three must be the same.)

You can also just look at pictures to see why this is wrong. Just draw a bunch of right triangles, "stand them up" on their long side (hypotenuse), and then draw the line straight down from the top point to the ground.

[RonPurewal]

In most cases"”including the one in this picture"”it should be clear that you're not getting 45º-45º-90º triangles.
Upon closer inspection, possibly including "mental rotations" of the small triangles to put them in the same position as the original triangle, you'll see that all three angles are the same as the three angles you started with.