Question

В.

Answer 1

Answer:

D). $\frac{3}{2} \pi\: units^2$

Explanation:

Let the $m\angle BOC = x\degree$

$\therefore m\angle AOB = 2x\degree$

$\because m\angle AOB+m\angle BOC = 180\degree\\..(straight \: line \: \angle 's) \\\therefore 2x + x = 180\degree \\\therefore 3x = 180\degree \\\\\therefore x = \frac{180\degree}{3}\\\\\therefore x = 60\degree \\\therefore 2x= 2\times 60\degree = 120\degree \\\implies m\angle BOC = 60\degree \\\therefore central \: \angle \: (\theta) = 60\degree \\Radius\: of \:circle \: (r) = 3\: units \\\\Area \: of\: shaded \: region\\\\ = \frac{\theta}{360\degree}\times \pi r^2 \\\\= \frac{60\degree}{360\degree}\times \pi \times 3^2 \\\\= \frac{1 }{6}\times \pi \times 9 \\\\= \frac{1 }{2}\times \pi \times 3 \\\\\huge \purple {\boxed {= \frac{3}{2} \pi\: units^2}}$

Answer 2

< aob+< boc = 180
< boc = x
2x + x = 180
X = 60
Area shaded = 60/360*#(3)^2 =3/2#