In a triangle ABC, I be the incenter and I1,I2 and I3 be the excenters. If R and r be the radius of circumcircle and incircle respectively, then II1.II2.II3=
A
R2r16
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4R2r
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
16R2r
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
16rR
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is C16R2r there's some standard result that says:
II1=4RsinA2
II2=4RsinB2
II3=4RsinC2
now II1.II2.II3=(4R)3[sinA2.sinB2.sinC2] -----equ(1)