Question

The sides of a triangle are in AP and its area is 3/5 x (area of an equilateral triangle of the same perimeter) Then, the ratio of the sides is

• A. 1:2:3
• B. 3:5:7
• C. 1:3:5
• D. None of these

Answer 1

The correct option is B

3:5:7

Here, 2b = a + c and $△=\frac{3}{5}\times \frac{\sqrt{3}}{4}{\left[\frac{a+b+c}{3}\right]}^2$
$\therefore △=\frac{\sqrt[3]{3[}3]}{20}{b}^2$
$\therefore sqrts(s-a)(s-b)(s-c)=\frac{\sqrt[3]{3[}3]}{20}.b^2$
or $\sqrt{\frac{1}{16}(a+b+c)(b+c-a)(c+a-b)(a+b-c)}=\frac{\sqrt[3]{33}}{20}{b}^2$
or $\sqrt{3b.(b+c+c-3b)b(2b-c+b-c)}=\frac{\sqrt[3]{33}}{20}{b}^2$
or $\sqrt{(2c-b)(3b-2c)}=\frac{3}{5}b$
or $8bc-4c^2-3b^2=\frac{9}{25}{b}^2$
or $\frac{84}{25}{b}^2-8bc+4c^2=0$
or $\frac{b}{c}=\frac{8\pm \sqrt{65-16\frac{84}{25}}}{2\frac{84}{25}=\frac{5}{7},\frac{5}{3}}$
also, 2b = a+c $\Rightarrow 2\frac{b}{c}=\frac{a}{c}+1$
$\Rightarrow \frac{a}{c}=\frac{2b}{c}-1=\frac{3}{7},\frac{7}{3}$