ExamHabit
Practise free

2000

JEE Main 2000 Mathematics question on Trigonometry

In a △ABC\triangle ABC, let ∠C=π/2\angle C = \pi/2. If rr is the inradius and RR is the circumradius of the triangle, then 2(r+R)2(r + R) is equal to
  1. A.
    a+ba + b
  2. B.
    b+cb + c
  3. C.
    c+ac + a
  4. D.
    a+b+ca + b + c
Show answer and solution

Correct option: A

Solution

For a right triangle with ∠C = π/2, the circumradius R = c/2 (hypotenuse is the diameter).

The inradius formula gives r = (a + b - c)/2.

So 2(r + R) = 2·[(a + b - c)/2 + c/2] = (a + b - c) + c = a + b.

[2000]

Practise Trigonometry the way it is asked

A timed set of real JEE Main Trigonometry questions, graded as the exam grades. No sign-up.

Start a Trigonometry test

More: all JEE Main Trigonometry questions · JEE Main 2000 · Mathematics chapter list