Let (5, a/4), be the circumcenter of a triangle with vertices A(a, −2), B(a, 6) and C(a/4, −2).

Question

Let \(\left(5, \frac{a}{4}\right)\), be the circumcenter of a triangle with vertices \(A(a,-2), B(a, 6)\) and \(C\left(\frac{a}{4},-2\right)\). Let \(\alpha\) denote the circumradius, \(\beta\) denote the area and \(\gamma\) denote the perimeter of the triangle. Then \(\alpha+\beta+\gamma\) is

(1) 60

(2) 53

(3) 62

(4) 30

Answer 1

Correct option is (2) 53

\(\mathrm{A}(\mathrm{a},-2), \mathrm{B}(\mathrm{a}, 6), \mathrm{C}\left(\frac{\mathrm{a}}{4},-2\right), \mathrm{O}\left(5, \frac{\mathrm{a}}{4}\right)\)

\(\mathrm{AO}=\mathrm{BO}\)

\((a-5)^{2}+\left(\frac{a}{4}+2\right)^{2}=(a-5)^{2}+\left(\frac{a}{4}-6\right)^{2}\)

\(\mathrm{a}=8\)

\(\mathrm{AB}=8, \mathrm{AC}=6, \mathrm{BC}=10\)

\(\alpha=5, \beta=24, \gamma=24\)