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

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

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?

Triangle QSR is the same as RSQ.

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 think the following post might helps: in-the-diagram-above-if-arc-abc-is-a-semicircle-what-is-119652.html#p1374777

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But 16+9 is the hypotenuse of the triangle PQR

Why are we using it as the height?

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.

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

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

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