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In Fig. 12.31, a square OABC is inscribe...

In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If `O A=20 c m ,` find the area of the shaded region. (Use `pi=3. 14`)

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Join OB.
We know `triangleOBA` is a right-angled triangle, as `angleOAB = 90^@ `(angle of a square)
Given, OA = `20` cm
Thus, OA = AB = `20` cm [Sides of a square]
Using Pythagoras theorem`OB^2 = OA^2 + AB^2`
= `(20 cm)^2 + (20 cm)^2`
OB = `(sqrt2) × (20 cm)^2`
= `20sqrt2` cm
Therefore, radius of the quadrant, r = OB = `20sqrt2` cm
Area of quadrant OPBQ =`((90^@)/(360^@)) xx pir^2`
= `1/4 xx 3.14 xx (20sqrt2 cm)^2`
= `1/4 xx 3.14 xx 400 xx 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`
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