(i)

Consider the above figure, Join PR,

Now PR = Diameter of the inscribed circle

Also, PR = BC = 10 cm.

So, PR = 10 cm

⇒ r = 5 cm

∴ Area of inscribed circle = πr² (put value of r in this equation)

⇒ Area of inscribed circle = π(5²)

⇒ Area of inscribed circle = 78.57 cm²

The area of inscribed circle is 78.57 cm².

(ii)

Consider the above figure, O is the centre of circle and ABCD is a square inscribed. Now OB and OD are radii of circle.

Consider ∆DBC right angled at c (as C is a vertex of square)

∴ Apply Pythagoras theorem in triangle DBC

Hypotenuse² = Perpendicular² + Base²

In triangle DBC, hypotenuse = DB,

perpendicular = BC and

base = DC

Put the values of BC and DC i.e. 10 cm

⟹ BD² = 200

⟹ BD = √200

⟹ BD = 10√2 cm

Now radius of circle = half of BD

⟹ r = (10√2)/2

⟹ r = 5√2 cm

Hence Area of circumscribing circle = πr²

⟹ Area of circumscribing circle = 3.14×5√2×5√2

(put π = 3.14 and r = 5√2 cm)

⇒ Area of circumscribing circle = 3.14 × 50

⇒ Area of circumscribing circle = 157 cm²

Area of circumscribing circle is 157 cm².