Question

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

• A. $3:4:5$
• B. $\sqrt{7}-1:\sqrt{7}:\sqrt{7}+1$
• C. $3:5:7$
• D. $2:1:8$

Answer 1

The correct option is C $3:5:7$

Let the sides be $a-d,a,a+d$
Perimeter $=3a\cdots \left(i\right)$.
$\therefore 2s=3a$
Area of triangle $\Delta =\sqrt{s(s-a^{\prime })(s-b^{\prime })(s-c^{\prime })}$
(where $a^{\prime },b^{\prime },c^{\prime }$ are sides of triangle)
Area of triangle $\Delta =\sqrt{\frac{3a}{2}(\frac{3a}{2}-a+d)(\frac{3a}{2}-a)(\frac{3a}{2}-a-d)}$
$\Delta =\sqrt{\frac{3a}{2}(\frac{a}{2}+d)\left(\frac{a}{2}\right)(\frac{a}{2}-d)}$
$\Delta =\sqrt{\frac{3a^2}{4}(\frac{a^2}{4}-d^2)}\cdots \left(i\right)$
Side of equilateral triangle having perimeter same as equation$\left(i\right)s^{\prime }=a$
Area of equilateral with same perimeter ${△}_E=\frac{\sqrt{3}{a}^2}{4}\cdots \left(ii\right)$
Area of given triangle=$\frac{3}{5}{\Delta }_E$
Taking square of both sides
${\Delta }^2=\frac{9}{25}{\Delta }_E^2$
$\Rightarrow \frac{3a^2}{4}(\frac{a^2}{4}-d^2)=\frac{9}{25}\times \frac{3a^4}{16}$
$\Rightarrow a^2-4d^2=\frac{9a^2}{25}\Rightarrow d=\frac{2a}{5}$
$\Rightarrow a-d:a:a+d=\frac{3a}{5}:a:\frac{7a}{5}=3:5:7$