Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
From a point within an equilateral triangle, perpendiculars drawn to the three sides are 6 cm, 7 cm, and 8 cm respectively. The length of the side of the triangle is
(a)7 cm
(b)10.5 cm
(c)$14 \sqrt{3} \mathrm{cm}$
(d)$\frac{14 \sqrt{3}}{3} \mathrm{cm}$
Answer
Answer (as printed): C
Explanation
Let each side of the triangle be $a \mathrm{cm}$. Then, $\operatorname{area}(\triangle \mathrm{AOB})+\operatorname{ar}(\triangle \mathrm{BOC})+\operatorname{ar}(\triangle \mathrm{AOC})=\operatorname{ar}(\triangle \mathrm{ABC})$ $$\begin{aligned} & \Rightarrow \frac{1}{2} \times a \times 6+\frac{1}{2} \times a \times 7+\frac{1}{2} \times a \times 8=\frac{\sqrt{3}}{4} a^{2} \\ & \Rightarrow \frac{a}{2}(6+7+8)=\frac{\sqrt{3}}{4} a^{2} \Rightarrow a=\left(\frac{21}{2} \times \frac{4}{\sqrt{3}}\right)=14 \sqrt{3} \mathrm{cm} . \end{aligned}$$