In given figure ABPC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region

Given: ABPC is a quadrant of circle of radius 14 cm.

BDCO is a semicircle, with BC as diameter.

We have

To find area of the shaded region, BDCP, we can write a relationship as

Area of shaded region, BDCP = Area of semicircle BDCO – (Area of quadrant ABPC – Area of ∆ABC) …(i)

We need to find each of these areas one by one.

Let’s start with area of quadrant ABPC. Area of quadrant ABPC is given by,

Area of quadrant ABPC = 1/4 (Area of circle)

= 1/4 πr²

= [∵ radius of quadrant is 14 cm; AB = AC = 14 cm]

= 154 cm² …(ii)

Now, we’ll find area of ∆ABC, which is given by

Area of ∆ABC = 1/2 × base of ∆ABC × height of ∆ABC

= 1/2 × AC × AB

= 1/2 × 14 × 14 [∵ AC = AB = 14 cm]

= 7 × 14 = 98 cm² …(iii)

To find area of semicircle BDCO, we need to find its radius BO or OC.

Using Pythagoras theorem in right triangle ABC, we can write

BC² = AB² + AC² [∵ (hypotenuse)² = (perpendicular)² + (base)²]

⇒ BC² = 14² +14² [∵ AB = AC = 14 cm]

⇒ BC² = 14² (1 + 1) [by taking 14² as common]

⇒ BC² = 14² × 2

⇒ BC = √(14² × 2)

⇒ BC = 14√2 cm

Clearly, BO = 1/2 BC

⇒ BO = 1/2 × 14√2

⇒ BO = 7√2 cm

Now, area of semicircle BDCO is given by,

Area of semicircle BDCO = 1/2 (area of circle)

= 1/2 (πr²)

= [∵ radius in semicircle BDCO = 7√2]

= 154 cm² …(iv)

Substituting equations (ii), (iii) and (iv) in equation (i), we get

Area of shaded region, BDCP = 154 – (154 – 98)

= 154 – 154 + 98

= 98 cm²

Hence, area of the shaded region = 98 cm²