Here is the basic mathematical way of solving this question.
Attachment:
File comment: Image related to solution of this problem
Traingle.jpg [ 9.23 KiB | Viewed 32302 times ]
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