Question

please solve this question guys​

Answer 1

Answer:

it may be 12 cm

but I am not sure

Answer 2

area of DPC=

pi/6*36=6picm^2

<br> area of

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

<br> Sector DP=

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

<br> Area of sector DP<br> PC

=2(3(2pi-3sqrt3

<br>

6(2pi-3sqrt3) cm^2

<br> Area of circulator part ADC

=pi/4*36=9pi

<br> Area of shaded region

=(9pi+9pi)-2(9sqrt3+6(2pi-3sqrt3)

<br>

18pi-2(9sqrt3+12pi-18sqrt3)

<br>

(18sqrt3-6pi) cm^2

<br> Shaded region=

6(3sqrt3-pi) cm^2