Question

CHAPTER NAME
VOLUME AND SURFACE AREA
PLZ HELP ME TO FIND ANS ​

Answer 1

$area \: of \: triangle \: abf = \frac{1}{2} (ab) \times (af) \times \sin(120) =$

$\frac{1} 2 \times {r}^{2} \times \frac{ \sqrt{3} }{2} =$

$\sqrt{3} \times {6}^{2} \div 4 = 9 \sqrt{3}$

segment BPF =sector - triangle

-=

$\frac{120}{360} \times \pi {r}^{2} - 9 \sqrt{3}$

$12\pi - 9\sqrt{3}$

area of shaded portion =hexagon-triangle

$6 \times \frac{ \sqrt{3} }{4} {r}^{2} - 9\sqrt{3}$

$54 \sqrt{3} - 9\sqrt{3} = 45 \sqrt{3}$