Multiple choice

In a triangle ABC, $I$ be the incentre and $I_1, I_2$ and $I_3$ be the excentres. If R and r be the radius of circum-circle and in-circle respectively. Then $I_1I_2=$

  1. $2R\sin{\dfrac{C}{2}}$
  2. $4R\sin{\dfrac{C}{2}}$
  3. $2R\cos{\dfrac{C}{2}}$
  4. $4R\cos{\dfrac{C}{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The distance between excentres I1 and I2 is given by the formula 4R*cos(C/2).

AI explanation

The excentre I1 is the intersection of the internal bisector of angle A and external bisectors of angles B and C, while I2 is the intersection of the internal bisector of angle B and external bisectors of A and C. In triangle I1CI2, the angle at C is 90 degrees minus C/2, making angle I1CI2 equal to 180 degrees minus 2 times 90 degrees minus C/2, which simplifies to C. Using the standard sine rule for triangle I1CI2, the side I1I2 equals 4R times the sine of angle I1CI2, where R is the circumradius of the original triangle ABC. Therefore, I1I2 equals 4R times the cosine of C/2.

Keep practicing — related questions