(a) 60 cm2
The radius ⊥ tangent at the pt. of contact,
therefore, OQ ⊥ PQ and OR ⊥ PR
∴ In rt. ΔΟPQ,
PQ = \(\sqrt{OP^2-OQ^2}\)
= \(\sqrt{169-25}\)
= \(\sqrt{144}\) = 12 cm
\(\Rightarrow\) PR = 12 cm (Two tangents from the same external pt. to a circle are equal)
Now area of quad. PQOR = 2 × Area of ΔPΟQ
= (2 x \(\frac{1}{2}\) x 12 x 5) cm2 = 60 cm2