A) 98.55 sq cm
B) 100 sq cm
C) 101 sq cm
D) 95 sq cm
Correct Answer: A
Solution :
[a] Radius of incircle\[=\frac{a}{2\sqrt{3}}=\frac{24}{2\sqrt{3}}=\frac{12}{\sqrt{3}}\] Area of triangle \[\frac{\sqrt{3}}{4}{{a}^{2}}=\frac{1.732}{4}\times {{(24)}^{2}}\] \[=\frac{1.732\times 576}{4}\] \[=249.408\] Area of circle \[=\pi {{r}^{2}}=\frac{22}{7}\times {{\left( \frac{12}{\sqrt{3}} \right)}^{2}}\] \[=\frac{22}{7}\times \frac{144}{3}=\frac{3168}{21}=150.857\] \[\therefore \]Area of remaining portion \[=249.408-150.857=98.55\,\,c{{m}^{2}}\] |
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