# 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.

Join OB.

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

Given, OA = 20 cm

Thus, OA = AB = 2o cm [Sides 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}

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 12

**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

**Summary:**

The area of the shaded region if a square OABC is inscribed in a quadrant OPBQ with OA = 20 cm is 228 cm^{2}.

**☛ Related Questions:**

- In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in Fig. 12.24. Find the area of the design.
- In Fig. 12.25, ABCD is a square of side 14 cm. With centres A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.
- Fig. 12.26 depicts a racing track whose left and right ends are semicircular. Fig. 12.26 The distance between the two inner parallel line segments is 60 m and they are each 106 m long. If the track is 10 m wide, find :(i) the distance around the track along its inner edge(ii) the area of the track.
- In Fig. 12.27, AB and CD are two diameters of a circle (with centre O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

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