Question

Choose the correct answer of the following question:

The height of an equilateral triangle is $3\sqrt{3}$ cm. Its area is

(a) $6\sqrt{3}$ cm² (b) 27 cm² (c) $9\sqrt{3}$ cm² (d) $27\sqrt{3}$ cm²

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{equilateral}\mathrm{triangle}\mathrm{be}x. \\ \mathrm{As},\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{equilateral}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times {\left(\mathrm{side}\right)}^2 \\ \Rightarrow \frac{1}{2}\times \mathrm{Base}\times \mathrm{Height}=\frac{\sqrt{3}}{4}\times x^2 \\ \Rightarrow \frac{1}{2}\times x\times 3\sqrt{3}=\frac{x^2\sqrt{3}}{4} \\ \Rightarrow \frac{3x\sqrt{3}}{2}=\frac{x^2\sqrt{3}}{4} \\ \Rightarrow \frac{3\sqrt{3}}{2}=\frac{x\sqrt{3}}{4} \\ \Rightarrow x=\frac{3\sqrt{3}\times 4}{2\sqrt{3}} \\ \Rightarrow x=6\mathrm{cm} \\ \mathrm{Now},\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}=\frac{\sqrt{3}}{4}\times 6^2 \\ =\frac{\sqrt{3}}{4}\times 36 \\ =9\sqrt{3}{\mathrm{cm}}^2 \end{gathered}$$

Hence, the correct answer is option (c).