Skip to main content
Back to Properties of Triangle
JEE Main 2024
Properties of Triangle
Properties of Triangle
Medium

Question

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

Options

Solution

A(a,−2),B(a,6),C(a4,−2),O(5,a4)AO=BO(a−5)2+(a4+2)2=(a−5)2+(a4−6)2a=8AB=8,AC=6,BC=10α=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}

Previous QuestionNext Question

Practice More Properties of Triangle Questions

View All Questions