Answer & Solution
Answer: Option A
Solution:
$$\eqalign{
& {\text{Area of half }}\Delta = \frac{1}{2}\left( {\frac{{\sqrt 3 }}{4}{a^2}} \right) \cr
& = \frac{1}{2} \times \frac{{\sqrt 3 }}{4} \times 6 \times 6 \cr
& = \frac{{9\sqrt 3 }}{2} \cr
& {\text{Area of shaded region}} \cr
& \Rightarrow \frac{{9\sqrt 3 }}{2} - \frac{{90}}{{360}}\pi {r^2} \cr
& = \frac{{9\sqrt 3 }}{2} - \frac{1}{4} \times \frac{{22}}{7} \times 1 \times 1 \cr
& = \frac{{9\sqrt 3 }}{2} - \frac{{11}}{{2 \times 7}} \cr
& = \frac{1}{2}\left( {9\sqrt 3 - \frac{{11}}{7}} \right){\text{c}}{{\text{m}}^2} \cr} $$
