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?

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72
96

108
150
200

The OA answer is 150

I treated this question as 2 45-45-90 Triangles and added the area

(We have the side 16 and 9 for each )

However the answer that I get is 132.5

ah, try again. these are DEFINITELY not 45-45-90 triangles, because to assume this would give two different values for the altitude. instead, consider that all three triangles in the diagram are similar to each other, and that should help you come up with a value for the altitude when you set up the appropriate ratios..

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Tim Sanders
Manhattan GMAT Instructor

Tim

Looks like I am missing out anything

This is how I looked at it Angle Q splits into 45 and 45

So Angle AQS has Q and A as 45 and S as 90 this is one of the Triangle

And the other is QST which has Q as 45 and S at 90 and T at 45 again

Could you please explain

Hi:

Here're my two cents on this.

Since, PQR is right angled at Q hence,

PR^2 = PQ^2 + QR^2

or 625 = PQ^2 + QR^2 ------- (1)

Since, angle(S) is 90 degrees, hence

PQ^2 = 16^2 + QS^2 and
QR^2 = 9^2 + QS^2

Put these values in equation (1):

QS = 12

Area of triange PQR = 1/2 * 12 * 25 = 150.

Hope this helps.

Angle Q does not split into two congruents angles. Assuming that angle Q splits into 45 and 45 is wrong.

To solve this kind of problem (there is one question very similar to this one in the MGMAT CATs), you can use two methods:

-You can use the pythagorean theorem as explained in the post above
-You can also use the similar triangles method that Tim mentionned

The similiar triangle method seems faster to me.


First, label the angle QPS 'x'.
Then label the angle QRS 'y'.

The big triangle QPR has angles of x°, y° and 90°.
So we know that x + y + 90 = 180

The triangle QPS has an angle of x° and an angle of 90°. So angle PQS is y°.

The triangle QSR has an angle of y° and an angle of 90°. So angle SQR is x°.



So, we have three similar triangles.

The sides of triangle PSQ are:
QS (side in front of angle x) : 16 (side in front of angle y) : PQ (side in front of angle 90)

The sides of triangle QRS are:
9 (side in front of angle x) : QS (side in front of angle y) : QR (side in front of angle 90)

So we can write the following equation:

QS/9 = 16/QS

Now, let's solve for QS:

QS² = 16*9
QS² = 144
QS = 12 (only the positive root makes sense for a side length)

With the altitude QS, we can now find the area:

(16+9)*12/2 = 25*12/2 = 300/2 = 150.

Excellent work David!

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Jamie Nelson
ManhattanGMAT Instructor

Thank you Jamie.

:-)

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Jamie Nelson
ManhattanGMAT Instructor