Learn

Saturday, 31 October

JEE Main 2024 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 29 January 2024, Shift 1, Maths Q14

Question 14 of 90 in this shift, Maths question 14 of 30, Section A.

Let (5,a4)\left(5,\frac{a}{4}\right) be the circumcenter of a triangle with vertices A(a,−2)A(a,-2), B(a,6)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
  1. (1)3030
  2. (2)6060
  3. (3)6262
  4. (4)5353Official answer

Official answer

Option 4

NTA final key.