ABCD is a square of side 6 cm. Find the area of the shaded region bounded between square and the two quadrants

Explanation

area of DPC=`pi/6*36=6picm^2`

area of`/_DPCC=sqrt3/4(36)=9sqrt3 cm^2`

Sector DP=`6pi-9sqrt3=3(2pi-3sqrt3) cm^2`

Area of sector DP

PC`=2(3(2pi-3sqrt3`

`6(2pi-3sqrt3) cm^2`

Area of circulator part ADC`=pi/4*36=9pi`

Area of shaded region`=(9pi+9pi)-2(9sqrt3+6(2pi-3sqrt3)`

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