# In Fig. 12.33, ABC 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

**Solution:**

We use the formula for areas of semi-circles and sector of circles to solve the problem.

To visualize the shaded region better, mark point D on arc BC of quadrant ABC and E on semicircle drawn with BC as diameter.

To find the area of semi-circle, we need to find the radius or diameter (BC) of the semicircle.

ΔABC is a right-angled Δ, right-angled at A (ABC being a quadrant).

Using Pythagoras theorem, we can find the hypotenuse (BC) of ΔABC.

ΔABC is a right angled Δ, right angled at A

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

= (14 cm)^{2} + (14 cm)^{2}

BC = √2 × (14 cm)²

= 14√2 cm

∴ Radius of semicircle BEC, r = BC/2 = 14√2/2 cm = 7√2 cm

Area of the shaded region = Area of semicircle - (Area of quadrant ABDC - Area ΔABC)

= πr^{2}/2 - (90°/360° × π(14)^{2} - 1/2 × AC × BC)

= π(7√2)^{2}/2 - (π(14)^{2}/4 - 1/2 × 1/4 × 1/4)

= [(22 × 7 × 7 × 2)/(7 × 2) - (22 × 14 × 14)/(7 × 4) - 7 × 14]

= 154 - (154 - 98)

= 98 cm^{2}

**Video Solution:**

## In Fig. 12.33, ABC 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

### NCERT Solutions Class 10 Maths - Chapter 12 Exercise 12.3 Question 15:

In Fig. 12.33, ABC 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

The area of the shaded region where ABC is a quadrant of a circle of radius 14 cm and a semicircle drawn with BC as diameter is 98 cm^{2}