Problem Solving

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?

I know the answer can be derived by using 3-4-5 triangle, my question is, how can you tell that triangle is 3-4-5 or 1-2-sqrt(3) with just those information given?
I mean is there an way to solve this without knowing that its 3-4-5 triangle?
or is there an way to just "know" that its 3-4-5 triangle by just looking at it?

I can definitely show how to solve this without knowing that it is a 3-4-5 triangle, but it is a little long
Let’s call the length PQ = a
Let’s call the length QR = b
Let’s call the length QS = c
We can say that 16^2 + c^2 = a^2
And that 9^2 + c^2 = b^2
Subtracting the two equations, we get
A^2 – b^2 = 175
From the larger triangle we also know that
A^2 + b^2 = 25^2
Again, adding these two equations will give us
A ^2 = 400
A = 20
From here it would be obvious to know that we have a 3 -4 -5, since sides will be 15, 20, 25
Hope I did that correctly. My answer would be D.
Again, there must be shorter and more efficient methods out there. Hopefully Ian and/or Stuart can also chip in.

D
use similar triangles
take triangles PQR and QSR
they both are similar triangles because 2 angles and a side are same

so,
QR/SR=PR/QR (base of PQR/base of QSR= hyp. of PQR/hyp of QSR)

just draw the triangles separately and you will get how I got he equation

so, QR/9=25/QR
QR=15
from this QS=sqrt(15^2-9^2)=12
area of tr PQR=1/2*25*12=150

yeah I am pondering over the same question, how do you tell just by looking at it that its a 3-4-5 triangle???

hongwang9703 wrote:yeah I am pondering over the same question, how do you tell just by looking at it that its a 3-4-5 triangle???

I usually go by this : if a right angle triangle has a hypotenuse as multiple of 5 then it is a 3-4-5 right triangle. I could be wrong but it solves most of the questions.

bump

my approach.

Cheers,

Hi ,

can some one explain this little briefly. Still i couldn't understand how we got the height. I thought i can do this by 45-45-90 triangle because the perpendicular divides the top angle by 2. Help greatly appreciated.

Andy.

mridul_dave wrote:

hongwang9703 wrote:yeah I am pondering over the same question, how do you tell just by looking at it that its a 3-4-5 triangle???

I usually go by this : if a right angle triangle has a hypotenuse as multiple of 5 then it is a 3-4-5 right triangle. I could be wrong but it solves most of the questions.

That's actually kind of an interesting observation. Obviously a right triangle with a hypotenuse of 5 needn't necessarily be a 3-4-5 triangle but the GMAT has a tendency to use integers for lengths and thus legs of 3 and 4 are the only integers that satisfy a right triangle of hypotenuse 5. Be careful not to get burned by that though!

anbuchand wrote:Hi ,

can some one explain this little briefly. Still i couldn't understand how we got the height. I thought i can do this by 45-45-90 triangle because the perpendicular divides the top angle by 2. Help greatly appreciated.

Andy.

a diff appraoch

in triangle PSQ and and PQR
angle psq=<pqr=90
<qpr=<spq
since two <s of two triangles are eq the 3rd will also be equal
hence tri PQR and PSQ are similar

similarly tri QSR~tri PQR--------> tri QSR ~ tri PSQ
------>PS/SQ=QS/SR
PS=16
SQ=9
--->qs^2=144--->qs=12

now in tri PQR base=25 ht=12
area=1/2 *12*25=150

This is what Cartman from south park means "I kick you in the nuts, Kenny!". There are some serious numbers involved here. I doubt you'll get a question like this on the gmat - maybe as a data sufficiency question, but not PS, specially not taking such big numbers.

Triangle PQR is a right triangle. QS is a height drawn through the right angle to the opposite side.

In any right triangle, a height drawn through the right angle to the opposite side creates 3 similar triangles.

We can prove this by plugging in for the angles measurements. If ∠RPQ = 20, then ∠PQS = 70, since the sum of the 2 angles must be 90. This forces ∠RQS to be 20. This forces ∠QRS to be 70. The result is that all 3 triangles (PQS, RQS, and PQR) have the same combination of angles: 20-70-90. Thus, the 3 triangles are similar.

The corresponding sides of similar triangles must yield the same proportion.
In triangle PQS, the shorter leg = QS, the longer leg = PS = 16.
In triangle RQS, the shorter leg = RS = 9, the longer leg = QS.
Since shorter leg:longer leg must be the same for each triangle, we get:
QS/16 = 9/QS
(QS)² = 144
QS = 12.

Thus, in PQR, h = 12 and b = PS + SR = 16+9 = 25.
Area = 1/2*b*h = 1/2(25)(12) = 150.

The correct answer is D.

Another approach:

In triangle PQR, b = PS + SR = 16+9 = 25.

Now we can plug in the answer choices, which represent the area of triangle PQR. Since b=25, the correct answer likely will be divisible by 25. Let's start with answer choice D.

Answer choice D: A = 150
1/2(25)(h) = 150
h = 12.
Now let's see if h=12 makes all the geometry work.
The legs of triangle PQS are 12 and 16, yielding a multiple of a 3:4:5 triangle. 3:4:5 = 12:16:20. Thus, PQ=20.
The legs of triangle RQS are 9 and 12, yielding another multiple of a 3:4:5 triangle. 3:4:5 = 9:12:15. Thus, QR = 15.
The legs of triangle PQR are 15 and 20 and the hypotenuse is 25, yielding another multiple of a 3:4:5 triangle. 3:4:5 = 15:20:25.
Since all the geometry works, success!

The correct answer is D.

zenithexe wrote: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?

I know the answer can be derived by using 3-4-5 triangle, my question is, how can you tell that triangle is 3-4-5 or 1-2-sqrt(3) with just those information given?
I mean is there an way to solve this without knowing that its 3-4-5 triangle?
or is there an way to just "know" that its 3-4-5 triangle by just looking at it?

Hello,

Since angle PQR = 90 and QS is perpendicular to PR I was just wondering if it is possible here that angle PQS = angle SQR = 45? Sorry if this is a trivial question. Thanks.

Best Regards,
Sri