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In Fig. 12.30, OACB is a quadrant of a c...

In Fig. 12.30, OACB is a quadrant of a circle with centre O and radius `3.5` cm. If `O D=2c m`, find the area of the
(i) quadrant OACB, (ii) shaded region.

Text Solution

Verified by Experts

Area of quadrant `OACB`, can be give as , `A = theta/360**pir^2`
Here, `theta = /_AOB = 90^@ and r = OB+OA = 2cm `
So, `A = 90/360**22/7**3.5**3.5 =1/4**11**3.5`
`A=38.5/4 = 9.625cm^2`

Now, Area of Shaded region,`A_S = ` Area of OACB - Area of `Delta OBD`
Area of `Delta OBD =1/2**OB**OD = 1/2**3.5**2 = 3.5cm^2`
So, area of shaded region, `A_S = 9.625 - 3.5 = 6.125cm^2`
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