Let be the circumcenter of a triangle with vertices and . Let denote the circumradius, denote the area, and denote the perimeter of the triangle. Then is

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Solution

Q.
Let $(5, \frac{a}{4})$ be the circumcenter of a triangle with vertices $A (a, - 2),$ $B (a, 6)$ and $C (\frac{a}{4}, - 2)$ . Let $α$ denote the circumradius, $β$ denote the area, and $γ$ denote the perimeter of the triangle. Then $α + β + γ$ is [2024]

1 62

2 60

3 30

4 53

Ans.
(4)

The circumcentre of the triangle ABC is $O (5, \frac{a}{4})$ $∵ O A^{2} = O B^{2}$

$\Rightarrow {(5 - a)}^{2} + {(\frac{a}{4} + 2)}^{2} = {(5 - a)}^{2} + {(\frac{a}{4} - 6)}^{2}$

$\Rightarrow 4 a = 32 \Rightarrow a = 8$

$∴$ $A \equiv (8, - 2), B \equiv (8, 6)$ and $C \equiv (2, - 2)$

$A C = \sqrt{36 + 0} = 6, B C = \sqrt{36 + 64} = 10, A B = \sqrt{64} = 8$

Perimeter, $γ = 6 + 10 + 8 = 24$

$Circumradius, α = O A = \sqrt{9 + 16} = 5$

$Area, β = \frac{1}{2} \times 6 \times 8 = 24$

$∴$ $α + β + γ = 5 + 24 + 24 = 53$

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