# In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

**Solution:**

We use the concepts of area of a sector of a circle and area of a square to solve the problem.

We can find the area of the quadrant OPBQ.

Join OB.

We know ΔOBA is a right-angled triangle, as ∠OAB = 90° (angle of a square)

∴ Using Pythagoras theorem

OB^{2} = OA^{2} + AB^{2}

= (20 cm)^{2} + (20 cm)^{2}

OB = √2 × (20 cm)²

= 20√2 cm

Therefore, radius of the quadrant, r = OB = 20√2 cm

Area of quadrant OPBQ = 90°/360° × πr^{2}

= 1/4 × 3.14 × (20√2 cm)^{2}

= 1/4 × 3.14 × 400 × 2 cm^{2}

= 628 cm^{2}

Area of square OABC = (side)^{2}

= (OA)^{2}

= (20 cm)^{2}

= 400 cm^{2}

Area of the shaded region = Area of quadrant OPBQ - Area of square OABC

= 628 cm^{2} - 400 cm^{2}

= 228 cm^{2}

**Video Solution:**

# In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region.

### NCERT Solutions Class 10 Maths - Chapter 12 Exercise 12.3 Question 13:

In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region

The area of the shaded region if a square OABC is inscribed in a quadrant OPBQ is 228 cm^{2}