Question

The sides of a triangle are in A.P. and its area is $\frac{3}{5}th$ of an equilateral triangle of same perimeter. Prove that its sides are in the ratio 3: 5: 7.

Answer 1

Let the sides be k -d, k, k+d i.e. a, b, c
$\therefore$ 2s = 3k $\therefore$ sides of equilateral $△$ is also k.
$S=\frac{3}{5}$ area of equilateral $△$ of same perimeter 3a
$S^2=\frac{9}{25}(\frac{\sqrt{3}}{4}{k}^2{)}^2$
$s(s-a)(s-b)(s-c)=\frac{9\times 3k^4}{25\times 16}$
$\frac{3k}{2}(\frac{3k}{2}-k+d)(\frac{3k}{2}-k)(\frac{3k}{2}-k)(\frac{3k}{2}-k-d)=\frac{27}{400}{k}^4$
$\frac{3k^2}{4}\frac{(k+2d)(k-2d)}{4}=\frac{27}{400}{k}^4$
or $k^2-4d^2=\frac{9}{25}{k}^2$ or $16k^2=100d^2$
$\therefore \frac{d}{k}=\sqrt{\frac{4}{25}}=\frac{2}{5}$
hence the sides k-d, k, k+d are in ratio
$1-\frac{d}{k},1,1+\frac{d}{k}$
or $1-\frac{2}{5},1,1+\frac{2}{5}$ or $3:5:7.$