Question

The sides of a triangle are in $A.P$ with middle term $b$ and common difference $a$. If its area is $\left(3/5\right)\times (area$ of an equilateral triangle of the same perimeter), then the area of the $\Delta$ is

• A. $\frac{\sqrt{3}}{20}{b}^2$
• B. $\frac{3\sqrt{3}}{20}{b}^2$
• C. $\frac{3\sqrt{3}}{20}(\frac{a+b}{3}{)}^2$
• D. $\frac{3\sqrt{3}}{20}{a}^2$

Answer 1

Answer: (B)

$\frac{3\sqrt{3}}{20}{b}^2$

The sides of the triangle are $b-a,b,b+a$.

Hence perimeter of the triangle is $P=3b$

Let the side of the equilateral triangle with same perimeter be s.

Therefore

$P_{equilateral}=P$

$3s=3b$
$\therefore s=b$.

Now, the area of the equilateral triangle

$=\frac{\sqrt{3}{s}^2}{4}$

$=\frac{\sqrt{3}{b}^2}{4}$.

Therefore the area of the given triangle

$=\frac{3}{5}\times \frac{\sqrt{3}{b}^2}{4}$

$=\frac{3\sqrt{3}{b}^2}{20}$