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

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:

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

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