Answer:
r = 9
Step-by-step explanation:
[tex] In\: \odot O, [/tex] QP is tangent at point P and OP is radius.
[tex] \therefore OP\perp QP[/tex]
[tex] \therefore m\angle OPQ = 90\degree [/tex]
So, [tex] \triangle OPQ[/tex] is a right angled triangle. Hence, OQ will be its HYPOTENUSE such that OQ = r + 32
Now, by Pythagoras theorem:
[tex] OQ^2 = OP^2 + PQ^2 [/tex]
[tex] \therefore (r + 32)^2 = r^2 + (40)^2 [/tex]
[tex] \therefore\cancel {r^2} + 64r + 1024 = \cancel {r^2} + 1600 [/tex]
[tex] \therefore 64r = 1600-1024 [/tex]
[tex] \therefore 64r = 576 [/tex]
[tex] \therefore r = \frac{576}{64}[/tex]
[tex] \therefore r = 9[/tex]
