Let left(5, frac{a}{4}right) be the circumcen | vedclass

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(B) Given vertices are $A(a, -2)$,$B(a, 6)$,and $C\left(\frac{a}{4}, -2\right)$.Since $A$ and $C$ have the same $y$-coordinate,$AC$ is a horizontal li

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$53$

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(B) Given vertices are $A(a, -2)$,$B(a, 6)$,and $C\left(\frac{a}{4}, -2ight)$.Since $A$ and $C$ have the same $y$-coordinate,$AC$ is a horizontal line segment of length $|a - \frac{a}{4}| = \frac{3a}{4}$.Since $A$ and $B$ have the same $x$-coordinate,$AB$ is a vertical line segment of length $|6 - (-2)| = 8$.Thus,$	riangle ABC$ is a right-angled triangle at $A$.The circumcenter of a right-angled triangle is the midpoint of the hypotenuse $BC$.The midpoint of $BC$ is $\left(\frac{a + a/4}{2}, \frac{6 - 2}{2}ight) = \left(\frac{5a}{8}, 2ight)$.Equating this to the given circumcenter $\left(5, \frac{a}{4}ight)$,we get $\frac{5a}{8} = 5 \implies a = 8$ and $\frac{a}{4} = 2$,which is consistent.With $a = 8$,the vertices are $A(8, -2)$,$B(8, 6)$,and $C(2, -2)$.Side lengths are $AB = 8$,$AC = |8 - 2| = 6$,and $BC = \sqrt{8^2 + 6^2} = 10$.Circumradius $\alpha = \frac{BC}{2} = \frac{10}{2} = 5$.Area $\beta = \frac{1}{2} 	imes AB 	imes AC = \frac{1}{2} 	imes 8 	imes 6 = 24$.Perimeter $\gamma = AB + AC + BC = 8 + 6 + 10 = 24$.Therefore,$\alpha + \beta + \gamma = 5 + 24 + 24 = 53$.

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

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