Question

If 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 sides are in the ratio

• A. $2:5:7$
• B. $3:5:7$
• C. $4:5:7$
• D. None of these

Answer 1

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

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 the triangle whose sides are in A.P.$=\sqrt{\frac{3a}{2}[\frac{3a}{2}-a+d][\frac{3a}{2}-a][\frac{3a}{2}-a-d]}$

$=\frac{\sqrt{3}a}{4}\sqrt{(a+2d)(a-2d)}$

${△}_1=\frac{\sqrt{3}a}{4}\sqrt{a^2-4d^2}$ ...(1)

$\because$ Perimeter of equilateral triangle $=$ Perimeter of the given triangle

$\therefore 3\times$one side of equilateral triangle $=3a$

$\Rightarrow$ side of the equilateral triangle $=a$

Now, ${△}_2=$Area of the 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, $\frac{\sqrt{a^2-4d^2}}{a}=\frac{3}{5}$

$25a^2-100d^2=9a^{2}16a^2=100d^2\Rightarrow \frac{a}{d}=\frac{5}{2}$

Ratio of the sides

$=a-d:a:a+d=\frac{a}{d}-1:\frac{a}{d}:\frac{a}{d}+1$

$=\frac{5}{2}-1:\frac{5}{2}:\frac{5}{2}+1=\frac{3}{2}:\frac{5}{2}:\frac{7}{2}=3:5:7.$