In triangle ABC,if B=90^{circ},then 2(r+R)= | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given,in $	riangle ABC$,$B=90^{\circ}$.Using the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R}$.Since $B=90^{\

Vedclass · Accepted Answer

$a+c$

Vedclass · Answer

(C) Given,in $	riangle ABC$,$B=90^{\circ}$.Using the sine rule,$\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c} = \frac{1}{2R}$.Since $B=90^{\circ}$,$\frac{\sin 90^{\circ}}{b} = \frac{1}{2R} \implies b = 2R$.The inradius $r$ is given by $r = (s-b) 	an(\frac{B}{2})$.Substituting $B=90^{\circ}$,$r = (s-b) 	an(45^{\circ}) = s-b$.Since $s = \frac{a+b+c}{2}$,we have $r = \frac{a+b+c}{2} - b = \frac{a-b+c}{2}$.Thus,$2r = a-b+c$.Finally,$2(r+R) = 2r + 2R = (a-b+c) + b = a+c$.

In $\triangle ABC$,if $B=90^{\circ}$,then $2(r+R)=$

Similar Questions

In triangle $ABC$,if $A=45^{\circ}$,$C=75^{\circ}$ and $R=\sqrt{2}$,then $r=$

In $\triangle ABC$,with usual notations,if $a \cos B = b \cos A$ and $a \cos C \neq c \cos A$,then the area of $\triangle ABC$ is . . . . . . sq. units.

If $P$ is a point on the altitude $AD$ of the $\triangle ABC$,and $\angle ABP = \frac{2B}{3}$,then $AP$ is equal to

If in triangle $ABC$,$\frac{a^2 - b^2}{a^2 + b^2} = \frac{\sin(A - B)}{\sin(A + B)}$,then the triangle is

In a triangle $ABC$ with usual notations,if $\tan A, \tan B, \tan C$ are in $H.P.$,then $a^{2}, b^{2}, c^{2}$ are in

Vedclass Test Series

Exam Paper Generator

Online Exam Module