# A chord with a length of 3  runs from pi/4  to pi/3  radians on a circle. What is the area of the circle?

Aug 2, 2016

$412.56$.

#### Explanation:

We use Co-ordinate Geometry to solve this Problem.

Without ant loss of generality, we may assume that the centre of the

circle is the origin.

A point $P , \frac{\pi}{4}$ radian on a circle of radius $r$ have co-ordinates

$P \left(r \cos \left(\frac{\pi}{4}\right) , r \sin \left(\frac{\pi}{4}\right)\right) = P \left(\frac{r}{\sqrt{2}} , \frac{r}{\sqrt{2}}\right)$.

Similarly, a point $Q , \frac{\pi}{3}$ radians on the circle is $Q \left(\frac{r}{2} , \frac{r \sqrt{3}}{2}\right)$.

Given that $P Q = 3 \Rightarrow P {Q}^{2} = 9$

$\Rightarrow {\left(\frac{r}{\sqrt{2}} - \frac{r}{2}\right)}^{2} + {\left(\frac{r}{\sqrt{2}} - \frac{r \sqrt{3}}{2}\right)}^{2} = 9$.

$\Rightarrow {\left(\frac{r \sqrt{2}}{2} - \frac{r}{2}\right)}^{2} + {\left(\frac{r \sqrt{2}}{2} - \frac{r \sqrt{3}}{2}\right)}^{2} = 9$.

$\Rightarrow {\left\{\frac{r}{2} \left(\sqrt{2} - 1\right)\right\}}^{2} + {\left\{\frac{r}{2} \left(\sqrt{2} - \sqrt{3}\right)\right\}}^{2} = 9$.

$\Rightarrow {r}^{2} / 4 \left\{\left(2 - 2 \sqrt{2} + 1\right) + \left(2 - 2 \sqrt{6} + 3\right)\right\} = 9$

$\Rightarrow {r}^{2} / 4 \left(8 - 2 \sqrt{2} - 2 \sqrt{6}\right) = 9$.

$\Rightarrow {r}^{2} = \frac{36}{8 - 2 \sqrt{2} - 2 \sqrt{6}}$

Taking, $\sqrt{2} \cong 1.414 , \sqrt{6} \cong 2.449$, we have,

${r}^{2} = \frac{36}{8 - 2.828 - 4.898} = \frac{36}{0.274} = 131.39$, so,

the area of the circle$= \pi {r}^{2} = 3.14 \cdot 131.39 = 412.56$