Let (5, frac a 4 ) , be the circumcenter of a triangle with vertices A(a,-2), B(a, 6) and C (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

Mathematics · properties of traingles

JEE Main 2024 — 29 January, Shift 1 — Question 8

Let $\left(5, \frac{\mathrm{a}}{4}\right)$ , 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

Option A:
60
Option B:
53
Correct
Option C:
62
Option D:
30

Answer: B

Step-by-step solution

$A(a,-2), B(a, 6), C\left(\frac{a}{4},-2\right), O\left(5, \frac{a}{4}\right)$

$\mathrm{AO}=\mathrm{BO}$

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

$\mathrm{a}=8$

$\mathrm{AB}=8, \mathrm{AC}=6, \mathrm{BC}=10$

$\alpha=5, \beta=24, \gamma=24$

$\alpha+\beta+\gamma=53$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 29 January, Shift 1
Subject: Mathematics
Chapter: properties of traingles
Topic: Sine,cosine,napier rules, half angle formula.area of triangle