Area

[shantascherla]

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?

Image -


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

[tim]

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..

[shantascherla]

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

[prakhar_au]

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.

[david.khoy]

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

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.

[jnelson0612]

Excellent work David!

[david.khoy]

Thank you Jamie.

[jnelson0612]

:-)

[jonvindjohnsen]

I recognized that the line PR equals to 25, which is a multiple of the 3:4:5 triangle.

So all the sides of PQR 15:20:25

And area= (15x20)/2 = 150..

Less than a min. During the test I just assumed that this would work out somehow.

But, is there any case that this will not work?


Thanks.

[RonPurewal]

I recognized that the line PR equals to 25, which is a multiple of the 3:4:5 triangle.

So all the sides of PQR 15:20:25

And area= (15x20)/2 = 150..

Less than a min. During the test I just assumed that this would work out somehow.

But, is there any case that this will not work?

sure, there are lots of cases in which it won't work.

* there are other right triangles with a hypotenuse of 25. there's even another one in all integers (7-24-25), in addition to the infinitude of such triangles with non-integer side lengths (e.g., 18-√301-25).

* and, of course, it won't work for triangles that are not right triangles.

on the other hand, on an exam like this one (i.e., an exam on which you aren't allowed to use a calculator), the 3-4-5 family is actually a pretty good guess.
especially if you don't have much in the way of other ideas about how to proceed.

[LuisN65]

I recognized that the line PR equals to 25, which is a multiple of the 3:4:5 triangle.

So all the sides of PQR 15:20:25

And area= (15x20)/2 = 150..

Less than a min. During the test I just assumed that this would work out somehow.

But, is there any case that this will not work?

sure, there are lots of cases in which it won't work.

* there are other right triangles with a hypotenuse of 25. there's even another one in all integers (7-24-25), in addition to the infinitude of such triangles with non-integer side lengths (e.g., 18-√301-25).

* and, of course, it won't work for triangles that are not right triangles.

on the other hand, on an exam like this one (i.e., an exam on which you aren't allowed to use a calculator), the 3-4-5 family is actually a pretty good guess.
especially if you don't have much in the way of other ideas about how to proceed.

Hi Ron,

One doubt: 3 - 4 - 5 right triangle is not a 30 - 60 - 90 right triangle? I am confused. I also answer the cat in the same way ---15:20:25

And area= (15x20)/2 = 150.. ---

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

[tim]

They are indeed different triangles. Did someone ever tell you otherwise? I'm curious as to why you would have even thought of this.

[RonPurewal]

One doubt: 3 - 4 - 5 right triangle is not a 30 - 60 - 90 right triangle?

they are totally different.

the ratio of the sides in a 30º-60º-90º triangle is 1 to √3 to 2. this ratio is THE ONLY USEFUL FACT about 30º-60º-90º triangles!.
literally, THE ONLY USEFUL FACT about them.
therefore, when you see a 30º-60º-90º triangle it should IMMEDIATELY come to mind... because it's the only thing that CAN come to mind. (apart from this, the only other thing you can really say about 30º-60º-90º triangles is ... that their angles are 30º, 60º, and 90º.)

the ratio 1:√3:2 is most certainly not the same as the ratio 3:4:5, so, there you go.

...but now you've made me curious—when you think '30º-60º-90º triangle', what association(s) have you been making?

[beakdas]

Hi Ron,

I was presented this question in my second MGMAT test. Now, looking at the solution it employs similarity of triangles as the modus operandi to resolve the problem presented above.

My question to you is that an assumption of pythagorean triplet with sides having the ratio 3:4:5 helped me decipher the problem swiftly. I am not sure how mathematically sound that approach is as obvious there could be cases where my assumption would not have been that profound.

Could you shed some light & help me understand why & how the selection of 'this' particular method worked?

Regards,
Abheek Das

[RonPurewal]

i'm not sure exactly what you are asking me.

if you're asking for the thought process behind this, then this is the process:
• 'hmm, this looks like it could be a 3-4-5 situation'
• 'let's see whether that works'
• 'hey, it works!'

that's it.
if it didn't work, that's when we would move on to trying more fancy things.