MathematicsMedium19×since 2002Q5031Let (5,a4)\left(5, \frac{a}{4}\right)(5,4a) be the circumcenter of a triangle with vertices A(a,−2),B(a,6)\mathrm{A}(a,-2), \mathrm{B}(a, 6)A(a,−2),B(a,6) and C(a4,−2)C\left(\frac{a}{4},-2\right)C(4a,−2). Let α\alphaα denote the circumradius, β\betaβ denote the area and γ\gammaγ denote the perimeter of the triangle. Then α+β+γ\alpha+\beta+\gammaα+β+γ isA60B62C53D30Check answerSkip