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.
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Dear Narenn,

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

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

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.

i had the same issue as in finding corresponding sides in similiar triangle. i suggest you watch some videos in khan academy relating to similiar triangles. there are few videos pertaining to it .watch them they are really helpful.

First of all, you don't show your work at all and asks what's wrong with it. Next, why do you assume that QSR is 45-45-90 triangle?
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But 16+9 is the hypotenuse of the triangle PQR

Why are we using it as the height?

Here is the basic mathematical way of solving this question.



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
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QS is not the median, it's height.
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