Straight Lines Question 7
Question 7 - 2024 (29 Jan Shift 1)
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
(1) 60
(2) 53
(3) 62
(4) 30
Show Answer
Answer (2)
Solution
$A(a,-2), B(a, 6), C\left(\frac{a}{4},-2\right), O\left(5, \frac{a}{4}\right)$
$A O=B O$
$(a-5)^{2}+\left(\frac{a}{4}+2\right)^{2}=(a-5)^{2}+\left(\frac{a}{4}-6\right)^{2}$
$a=8$
$A B=8, A C=6, B C=10$
$\alpha=5, \beta=24, \gamma=24$
