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
$3^{\text{rd }}\Delta =3h\text{ etc. }$
now side of the $1^{\text{st }}\Delta$ is ' $a^{\prime }=\frac{2h}{\sqrt{3}}$
$\therefore a_2=\frac{2}{\sqrt{3}}\cdot \sqrt{3}h=2h$
$a_3=2\sqrt{3}h$
$\text{ Mly }{a}_6=18h$
$\text{ now }\frac{\sqrt{3}}{4}\cdot (18h{)}^2-\frac{\sqrt{3}}{4}\cdot \frac{4h^2}{3}=968\sqrt{3}$
$81h^2-\frac{h^2}{3}=968\Rightarrow \frac{242h^2}{3}=968\Rightarrow h^2=12\Rightarrow h=2\sqrt{3}$
$\therefore a=\frac{2}{\sqrt{3}}\cdot 2\sqrt{3}=4$
$\text{ perimeter }=12$