# A triangle has corners at (5 ,8 ), (2 ,6 ), and (7 ,3 ). What is the area of the triangle's circumscribed circle?

##### 1 Answer
Aug 11, 2018

The area of the triangle's circumscribed circle is:
$\Delta \approx 23.2499 , s q . u n i t s$

#### Explanation:

Let , $\triangle A B C$ be the triangle with corners at

$A \left(5 , 8\right) , B \left(2 , 6\right) \mathmr{and} C \left(7 , 3\right) .$

Using Distance formula ,we get

$a = B C = \sqrt{{\left(7 - 2\right)}^{2} + {\left(3 - 6\right)}^{2}} = \sqrt{25 + 9} = \sqrt{34}$

$b = C A = \sqrt{{\left(5 - 7\right)}^{2} + {\left(8 - 3\right)}^{2}} = \sqrt{4 + 25} = \sqrt{29}$

$c = A B = \sqrt{{\left(5 - 2\right)}^{2} + {\left(8 - 6\right)}^{2}} = \sqrt{9 + 4} = \sqrt{13}$

Using cosine Formula ,we get

$\cos B = \frac{{c}^{2} + {a}^{2} - {b}^{2}}{2 c a} = \frac{13 + 34 - 29}{2 \sqrt{13} \sqrt{34}} = \frac{18}{2 \sqrt{442}} = \frac{9}{\sqrt{442}}$

We know that,

${\sin}^{2} B = 1 - {\cos}^{2} B$

$\implies {\sin}^{2} B = 1 - \frac{9}{442} = \frac{433}{442}$

$\implies \sin B = \left(\sqrt{\frac{433}{442}}\right) \to \left[\because B \in \left({0}^{\circ} , {180}^{\circ}\right)\right]$

Using sine formula:we get

$\frac{b}{\sin} B = 2 R \implies R = \frac{b}{2 \sin B}$

$\implies R = \frac{\sqrt{29}}{2 \left(\sqrt{\frac{433}{442}}\right)} \approx 2.75$

So , the area of the triangle's circumscribed circle is:

$\Delta = \pi {R}^{2} = \pi \cdot {\left(\frac{\sqrt{29}}{2 \left(\sqrt{\frac{433}{442}}\right)}\right)}^{2} = \pi \left(\frac{29 \times 442}{4 \times 433}\right)$

$\Delta \approx 23.2499 , s q . u n i t s$