Question

Thea area of an equilateral triangle inscribed in the circle $x^2+y^2-2x=0$ is:

• A. $\frac{3\sqrt{3}}{2}$
• B. $\frac{3\sqrt{3}}{4}$
• C. $\frac{3\sqrt{3}}{8}$
• D. $Noneofthese$

Answer 1

The correct option is B $\frac{3\sqrt{3}}{4}$

We have general equation of a circle as,

$(x-a{)}^2+(y-b{)}^2=r^2$, The given equation can be rewritten as follows,

$(x-1{)}^2+y^2=1$ , here we have radius `r' of circle as 1. Consider the equilateral triangle $△ABC$ inscribed in the circle as given in the figure. From the right angled triangle ADO, we calculate the distance OD as $AOsin{30}^{\circ }$ and the distance AD is calculated as $AOcos{30}^{\circ }$.

The area of the triangle = $\frac{1}{2}bh$

$=\frac{1}{2}AB\times CD$

we have $AB=2AD$

which is equal to $\sqrt{3}$ and $CD=OC+OD=r+OD$ is equal to $\frac{3}{2}$. And the area of the equilateral triangle inscribed in the circle is

Area =$\frac{1}{2}\sqrt{3}\times \frac{3}{2}$$=\frac{3\sqrt{3}}{4}$