JEE MAIN - Mathematics (2024 - 29th January Morning Shift - No. 3)
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
60
62
53
30
Explanation
$$\begin{aligned}
& A(a,-2), B(a, 6), C\left(\frac{a}{4},-2\right), O\left(5, \frac{a}{4}\right) \\
& A O=B O \\
& (a-5)^2+\left(\frac{a}{4}+2\right)^2=(a-5)^2+\left(\frac{a}{4}-6\right)^2 \\
& a=8 \\
& A B=8, A C=6, B C=10 \\
& \alpha=5, \beta=24, \gamma=24
\end{aligned}$$
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