Question

Area of the equilateral triangle inscribed in the circle $x^2+y^2-7x+9y+5=0$ is

• A. $\frac{155}{8}\sqrt{3}squnit$
• B. $\frac{165}{8}\sqrt{3}squnit$
• C. $\frac{175}{8}\sqrt{3}squnit$
• D. $\frac{185}{8}\sqrt{3}squnit$
• E. $\frac{195}{8}\sqrt{3}squnit$

Answer 1

Answer: (B)

Given, $x^2-y^2-7x+9y+5=0$
$\therefore$ $R=\sqrt{{\left(\frac{-7}{2}\right)}^2+{\left(\frac{9}{2}\right)}^2-5}$
$=\sqrt{\frac{49}{4}+\frac{81}{4}-5}=\frac{\sqrt{110}}{2}$
In $\Delta OAB,$ $\cos 30{}^{\circ }=\frac{AB}{R}$
$\Rightarrow$ $\frac{\sqrt{3}}{2}=\frac{AB}{\sqrt{110/2}}$

$\Rightarrow$ $\frac{\sqrt{330}}{4}=AB$
$\therefore$ Length $AC=\frac{\sqrt{330}}{2}$
$\therefore$ Area of equilateral $\Delta =\frac{\sqrt{3}}{4}{(a)}^2$
$=\frac{\sqrt{3}}{4}\times \frac{330}{4}=\frac{165\sqrt{3}}{8}$ sq unit