A) \[48\sqrt{3}c{{m}^{2}}\]
B) \[16\sqrt{3}c{{m}^{2}}\]
C) \[15\sqrt{3}c{{m}^{2}}\]
D) \[20\sqrt{3}c{{m}^{2}}\]
Correct Answer: A
Solution :
Let each side of the triangle be a. Then, Its height \[=\frac{\sqrt{3}}{2}\times a\] \[\therefore \]\[\frac{\sqrt{3}}{2}a=12\,\,cm\] \[a=\frac{12\times 2}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{24\sqrt{3}}{3}=8\sqrt{3}cm\] Side \[=8\sqrt{3}cm\] \[\therefore \]area \[=\frac{\sqrt{3}}{4}\times {{\left( a \right)}^{2}}\] sq. Units \[=\frac{\sqrt{3}}{4}\times {{\left( 8\sqrt{3} \right)}^{2}}=\frac{\sqrt{3}}{4}\times 64\times 3=48\sqrt{3}c{{m}^{2}}\]
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