JEE Math Practice Question
Let $$\left(5, \frac{a}{4}\right)$$ be the circumcenter of a triangle with vertices $$\mathrm{A}(a,-2), \mathrm{B}(a, 6)$$ 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
- A.60
- B.62
- C.53
- D.30
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