# Easy Geometry & Mensuration Solved QuestionAptitude Discussion

 Q. The perimeter of an equilateral $\triangle$ is $72\sqrt{3}$ meters. Find its height.
 ✖ A. 63 metres ✖ B. 55 metres ✖ C. 40 metres ✔ D. 36 metres

Solution:
Option(D) is correct

Let one side of the $\triangle$ be $=a$

Perimeter of equilateral $\triangle=3a$

$\Rightarrow 3a=72\sqrt{3}$

$\Rightarrow a=24\sqrt{3}$

Height $=AC$ by pythagoras theorem

$AC^2=a^2-\left(\dfrac{a}{2}\right)^2$

$AC^2=a^2\times \left[1-\left(\dfrac{1}{2}\right)^2\right]$

$AC^2=a^2\times \left[1- \dfrac{1}{4}\right]$

$AC^2=a^2\times \dfrac{3}{4}$

Now, putting, $a=24\sqrt{3}$

$\Rightarrow AC^2 = 24^2 \times 3 \times \dfrac{3}{4}$

$\Rightarrow AC^2 = \dfrac{24^2\times 3^2}{2^2}$

$\Rightarrow AC = \dfrac{24\times 3}{2}$

$AC=36\text{ cm}$

Edit: For an alternative solution with less calculation, check comment from Thulasi.

## (3) Comment(s)

Thulasi
()

$3a=72\sqrt{3}$

$\Rightarrow a=24\sqrt{3}$

Now, area of an equilateral triangle, $\dfrac{3}{4} \times (\text{side})^2$ -------- (1)

Also, we know that area of any triangle is, $\dfrac{1}{2} \times \text{base} \times \text{height}$ -------- (2)

From (1) and (2),

$\Rightarrow \dfrac{3}{4}\times a^2 = \dfrac{1}{2} \times a \times h$

$\dfrac{3}{4} \times 24 \sqrt{3} = \dfrac{1}{2} \times h$

$h = \textbf{36 cms}$

KARTIK
()