Verified Solution • 2025–26 Session moderate

QJEEM24Ma3c23: JEE Main 2024 Mathematics - Class 11 JEEM24Ma3c23 Solution: Let be the circumcenter of a triangle with vertices and . Let…

Let (5,a4)\left(5, \frac{a}{4}\right) be the circumcenter of a triangle with vertices A(a,−2),B(a,6)\mathrm{A}(a,-2), \mathrm{B}(a, 6) and C(a4,−2)C\left(\frac{a}{4},-2\right) . Let α\alpha denote the circumradius, β\beta denote the area and γ\gamma denote the perimeter of the triangle. Then α+β+γ\alpha+\beta+\gamma is

Multiple Choice Options

  • A 60
  • B 62
  • C 53
  • D 30

Step-by-Step Verified Solution

amp;A(a,−2),B(a,6),C(a4,−2),O(5,a4)amp;AO=BOamp;(a−5)2+(a4+2)2=(a−5)2+(a4−6)2amp;a=8amp;AB=8,AC=6,BC=10amp;α=5,β=24,γ=24\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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