In the diagram to the right, triangle PQR has a right angle

Question

https://gmatclub.com/forum/download/file.php?id=22800 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? A. 72 B. 96 C. 108 D. 150 E. 200 Capture.PNG

Bunuel · Accepted Answer

Important property: the perpendicular to the hypotenuse will always divide the triangle into two triangles with the same properties as the original triangle. Check here: if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html#p1154669 According to the above, triangles PQR, PSQ and QSR must be similar. Now, since triangles PSQ and QSR are similar, the the ratio of their corresponding sides must be the same (corresponding sides are opposite equal angles): PS/QS = QS/SR --> QS^2 = PS*SR = 16*9 --> QS = 4*3 = 12. The area of triangle PQR = 1/2*base*height = 1/2*(16+9)*12 = 150. Answer: D. Hope it's clear.

Narenn · Answer

It's a standard relationship based on similar triangle theory.

We should note that triangle PQR, triangle PSQ, and triangle QSR are similar

In similar triangles, ratios of corresponding sides are equal.

We know that triangle PSQ is similar to triangle QSR

so,  \frac{base of PSQ}{height of PSQ} = \frac{base of QSR}{height of QSR}

\frac{QS}{PS} = \frac{RS}{QS}  -------->  QS^2  = PS*RS

Similarly you can also prove that  PQ^2 = PS*PR and  QR^2  = SR*PR

Narenn · Answer

In right triangle PQR right angled at Q, QS^2 = PS * SR --------> qs^2 = 16*9 ------> qs = 4*3 = 12

Area =  \frac{1}{2} QS*PR  ----->  \frac{1}{2} 12*25  ------> 150

sudeeptasahu29 · Answer

Dear Narenn,

Thank you for the reply. Please explain how you got this relation QS^2 = PS * SR??

Kconfused · Answer

Hello Bunnel, how can we ascertain that PSQ and QSR but not PSQ and RSQ that are similar?

Bunuel · Answer

Triangle QSR is the same as RSQ.

Kconfused · Answer

Thanks Bunnel!
Wouldn't that mix up the corresponding sides? Obviously QSR and RSQ when matched with PSQ would result in different set of corresponding sides
My question is, how can we find the right set of corresponding sides to match up in similar triangles?
Thank you in advance! I've always had this problem with similar triangles.

Bunuel · Answer

I think the following post might helps: in-the-diagram-above-if-arc-abc-is-a-semicircle-what-is-119652.html#p1374777 Similar questions to practice: in-the-diagram-to-the-right-triangle-pqr-has-a-right-angle-169506.html in-the-figure-above-the-circles-are-centered-at-o-0-0-and-162390.html in-the-figure-above-ad-4-ab-3-and-cd-168366.html in-the-figure-above-segments-rs-and-tu-represent-two-positi-167496.html in-triangle-pqr-the-angle-q-9-pq-6-cm-qr-8-cm-x-is-161440.html in-triangle-abc-to-the-right-if-bc-3-and-ac-4-then-88061.html if-arc-pqr-above-is-a-semicircle-what-is-the-length-of-144057.html in-triangle-abc-if-bc-3-and-ac-4-then-what-is-the-126937.html what-are-the-lengths-of-sides-no-and-op-in-triangle-nop-130657.html in-the-diagram-what-is-the-length-of-ab-127098.html length-of-ac-119652.html in-the-above-circle-ab-4-bc-6-ac-5-and-ad-6-what-106009.html triangle-pqr-is-right-angled-at-q-qt-is-perpendicular-to-pr-106177.html

zanaik89 · Answer

But 16+9 is the hypotenuse of the triangle PQR

Why are we using it as the height?

Bunuel · Answer

We are considering PR = 25 as the base of triangle PQR and QS as the height:

The area of triangle PQR = 1/2*base*height = 1/2*PR*QS = 1/2*(16+9)*12 = 150.

GMATisLovE · Answer

Here is the basic mathematical way of solving this question.

Attachment:

File comment: Image related to solution of this problem

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Let QR = x, PQ = y and QS = z
Given PS = 16 and SR = 9

Triangle PQR is right angle triangle
\(x^2 + y^2 = (16+9)^2 = 25^2\) -----------------------------------------Equation 1

Triangle QSR is right angle triangle

\(z^2 + 9^2 = x^2\) ------------------------------------------------Equation 2

Triangle QPS is right angle traingle

\(z^2 + 16^2 = y^2\) -----------------------------------------------Equation 3

Putting the values of x^2 and y^2 from equation 2 and 3 into equation 1, we get

\(z^2 + 9^2 + z^2 + 16^2 = 25^2\)

\(2z^2 + 9^2 + 16^2 = 25^2\) ---------------------------------Equation 4

Solving equation 4 for z, we get z=12

Now Area of the triangle is (\(\frac{1}{2}\))*base*hieght
So, area of triangle PQR is (\(\frac{1}{2}\))*PR*QS = (\(\frac{1}{2}\))*(16+9)*12 = (\(\frac{1}{2}\))*25*12 = 150

Hence D

Sherlock221b · Answer

Standard type of right triangles are with sides 3 4 5 , 5 12 13, 8 15 17, and 7 24 25. 1) In this question we have hypotenuse as 25 (a multiple of 5) so we can choose 3 4 5 and 5 12 13. 2) Now hypotenuse is the largest side of a right triangle , this leaves only 3 4 5. In this case multiply each side of 3 4 5 with 5 (since hypotenuse is a multiple of 5) we get 15 20 25. Now calculate the area! (15*20)/2=150 hit kudos if you like or tell me where I am wrong

bumpbot · Answer

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In the diagram to the right, triangle PQR has a right angle

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In the diagram to the right, triangle PQR has a right angle

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