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In fig. 3, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

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#### Solution

Let us join OB.

In ΔOAB:

OB^{2} = OA^{2} + AB^{2} = (20)^{2} + (20)^{2} = 2 × (20)^{2}

⇒ OB = 20 √2

Radius of the circle, *r *= `20 sqrt2` cm

`"Area of qudrant OBPQ"=90^@/360^@xx3.14xx(20sqrt2)^2`

`=90/360xx3.14xx(20sqrt2)^2 cm^2`

`=1/4xx3.14xx800 cm^2`

`=628 cm^2`

Area of square OABC = (Side)^{2} = (20)^{2} cm^{2} = 400 cm^{2}

∴ Area of the shaded region = Area of quadrant OPBQ − Area of square OA

= (628 − 400) cm^{2}

= 228 cm^{2}

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