Question

A sequence of equilateral triangles is drawn. The altitude of each is $\sqrt{3}$ times the altitude of the preceding triangle, the difference between the area of the first triangle and the sixth triangle is $968 \sqrt{3}$ square unit. The perimeter of the first triangle is

• A. 10
• B. 12
• C. 16
• D. 18

Answer 1

Answer: (B)

Let the altitude of the $1^{\text {st }} \Delta= h$
altitude of $2^{\text {nd }} \Delta=\sqrt{3}$ h
$3^{\text {rd }} \Delta=3 h \text { etc. }$
now side of the $1^{\text {st }} \Delta$ is ' $a ^{\prime}=\frac{2 h }{\sqrt{3}}$
$\therefore a _2=\frac{2}{\sqrt{3}} \cdot \sqrt{3} h =2 h $
$a _3=2 \sqrt{3} h$
$\text { Mly } a _6=18 h $
$\text { now } \frac{\sqrt{3}}{4} \cdot(18 h )^2-\frac{\sqrt{3}}{4} \cdot \frac{4 h ^2}{3}=968 \sqrt{3}$
$ 81 h ^2-\frac{ h ^2}{3}=968 \Rightarrow \frac{242 h ^2}{3}=968 \Rightarrow h ^2=12 \Rightarrow h =2 \sqrt{3}$
$\therefore a =\frac{2}{\sqrt{3}} \cdot 2 \sqrt{3}=4$
$ \text { perimeter }=12$