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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)
We use the concepts of area of a sector of a circle and area of a square to solve the problem.
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
OB2 = OA2 + AB2
= (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° × πr2
= 1/4 × 3.14 × (20√2 cm)2
= 1/4 × 3.14 × 400 × 2 cm2
= 628 cm2
Area of square OABC = (side)2
= (20 cm)2
= 400 cm2
Area of the shaded region = Area of quadrant OPBQ - Area of square OABC
= 628 cm2 - 400 cm2
= 228 cm2
☛ Check: NCERT Solutions for Class 10 Maths Chapter 12
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
The area of the shaded region if a square OABC is inscribed in a quadrant OPBQ with OA = 20 cm is 228 cm2.
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