Question

The sides of a triangle are in A.P. and its area is $\frac{3}{5}$th of an equilateral triangle of the same perimeter, then the ratio of its sides

Answer 1

Let the sides be $a-d,a$ and $a+d$

$2s=$ sum of the sides $=3a\therefore s=\frac{3a}{2}$
Now, ${△}_1=$ Area of triangle whose sides are in $A.P.$
$=\sqrt{\frac{3a}{2}\left[\frac{3a}{2}-a+d\right]\left[\frac{3a}{2}-a\right]\left[\frac{3a}{2}-a-d\right]}$ $=\frac{3a}{4}\sqrt{(a+2d)(a-2d)}$
or, ${△}_1=\frac{3a}{4}\sqrt{a^2-4d^2}$ ...(1)
$\because$ Parameter of equilateral triangle $=$ Parameter of the given triangle
$\therefore$ $3\times$ one side of equilateral triangle $=3a$ $\Rightarrow$ side of the equilateral triangle $=a$
Now, ${△}_2=$ Area of equilateral triangle
$=\frac{\sqrt{3}}{4}\times {\left(side\right)}^2=\frac{\sqrt{3}}{4}{a}^2$ ...(2)
From equation, $\frac{{△}_1}{{△}_2}=\frac{3}{5}$ or $\frac{\sqrt{a^2-4d^2}}{a}=\frac{3}{5}$ or, ${25}^2-{100d}^2=9a^2$ or,$16a^2=100d^2\Rightarrow \frac{a}{d}=\frac{5}{2}$
Ratio of the sides
$=a-d:a:a+d=\frac{a}{d}-1:\frac{a}{d}:\frac{a}{d}+1$ $=\frac{5}{2}-1:\frac{5}{2}:\frac{5}{2}+1=\frac{3}{2}:\frac{5}{2}:\frac{7}{2}=3:5:7$