Subject: General Aptitude · Chapter: Area · Exam: 2008 · Marks: · Difficulty:
The altitude of an equilateral triangle of side $2 \sqrt{3} \mathrm{cm}$ is
(a)$\frac{1}{2} \mathrm{cm}$
(b)$\frac{\sqrt{3}}{4} \mathrm{cm}$
(c)$\frac{\sqrt{3}}{2} \mathrm{cm}$
(d)3 cm
Answer
Answer (as printed): D
Explanation
Let ABC be the equilateral triangle and AD be the altitude on base BC. In an equilateral triangle, the altitude and the median coincide. So, $\mathrm{BC}=\mathrm{DC}=\left(\frac{2 \sqrt{3}}{2}\right) \mathrm{cm}=\sqrt{3} \mathrm{cm}$. Let the length of the altitude AD be $x \mathrm{cm}$. Then, in right angled $\triangle \mathrm{ADB}$, $\mathrm{AB}^{2}=\mathrm{AD}^{2}+\mathrm{BD}^{2} \Rightarrow(2 \sqrt{3})^{2}=x^{2}+(\sqrt{3})^{2} \Rightarrow x^{2}=(12-3)$ $=9 \Rightarrow x=3 \mathrm{cm}$.