Radius of the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in ΔABC,

BC^{2 }= AB^{2 }+AC^{2}

⇒ BC^{2 }= 14^{2 }+14^{2}

⇒ BC = \( 14\sqrt{2}\) cm

Radius of semicircle = \( \dfrac{14\sqrt{2}}{2} \) cm = 7\( \sqrt{2}\) cm

Area of ΔABC = (\( \dfrac{1}{2}\))×14×14 = 98 cm^{2}

Area of quadrant = \( (\dfrac{1}{4})×(\dfrac{22}{7})×(14×14)\) = 154 cm^{2}

Area of the semicircle = \( (\dfrac{1}{2})×(\dfrac{22}{7})×7\sqrt{2}×7\sqrt{2}\) = 154 cm^{2}

Area of the shaded region =Area of the semicircle + Area of ΔABC – Area of quadrant

= 154 +98-154 cm^{2 }= 98cm^{2}

Prove that the area of a circular path of uniform width hsurrounding a circular region of radius r is πh(2r + h).

A road which is 7 m wide surrounds a circular park whose circumference is 352 m. Find the area of the road.

The outer circumference of a circular race-track is 528 m. The track is everywhere 14 m wide. Calculate the cost of levelling the track at the rate of 50 paise per square metre

A circular pond is of diameter 17.5 m. It is surrounded by a 2 m wide path. Find the cost of constructing the path at the rate of ₹25 per square metre (Use π = 3.14)

A path of width 3.5 m runs around a semicircular grassy plot whose perimeter is 72 m. Find the area of the path.