You visited us 1 times! Enjoying our articles? Unlock Full Access!

Relation between Inradius and Perimeter of Triangle

In Δ ABC ...

Question

In $Δ$ $A B C$ the sides opposite to angles $A, B, C$ are denoted by $a, b, c$ respectively.
If $I$ is the incenter of a triangle $A B C$ , then the ratio $I A : I B : I C$ is equal to?

c o s e c \frac{A}{2} : c o s e c \frac{B}{2} : c o s e c \frac{C}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

s i n \frac{A}{2} : s i n \frac{B}{2} : s i n \frac{C}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

s e c \frac{A}{2} : s e c \frac{B}{2} : s e c \frac{C}{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of the above

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $c o s e c \frac{A}{2} : c o s e c \frac{B}{2} : c o s e c \frac{C}{2}$
$I A = \frac{r}{sin (\frac{A}{2})}, I B = \frac{r}{sin (\frac{B}{2})}, I C = \frac{r}{sin (\frac{C}{2})} I A : I B : I C = \frac{r}{sin (\frac{A}{2})} : \frac{r}{sin (\frac{B}{2})} : \frac{r}{sin (\frac{C}{2})} I A : I B : I C = c o s e c (\frac{A}{2}) : c o s e c (\frac{A}{2}) : c o s e c (\frac{A}{2})$

Suggest Corrections

Similar questions

Δ

A B C

A, B, C

a, b, c

cos A + sin A - \frac{2}{cos B + sin B} = 0,

{(\frac{a + b}{c})}^{4}

Δ

A B C

A, B, C

a, b, c

\frac{s i n A}{4} = \frac{s i n B}{5} = \frac{s i n C}{6}

c o s A + c o s B + c o s C

Δ

A B C

A, B, C

a, b, c

∠

π / 2

tan (C / 2)

Δ

A B C

A, B, C

a, b, c

3 a = b + c

tan \frac{B}{2} \times tan \frac{C}{2}

Δ

A B C

A, B, C

a, b, c

(sin A + sin B + sin C) (sin A + sin C - sin B) = μ sin A sin C

sin A = a k, sin B = b k, sin C = c k

μ

Join BYJU'S Learning Program