In triangle ABC,suppose the radius of the exc | vedclass

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

Question

(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area and $s$ is the semi-p

Vedclass · Accepted Answer

$r$

Vedclass · Answer

(B) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$,where $\Delta$ is the area and $s$ is the semi-perimeter of the triangle.Then,$ab - r_1 r_2 = ab - \frac{\Delta^2}{(s-a)(s-b)}$.Since $\Delta^2 = s(s-a)(s-b)(s-c)$,we have $ab - r_1 r_2 = ab - s(s-c)$.Substituting $s = \frac{a+b+c}{2}$,we get $ab - \frac{a+b+c}{2} \cdot \frac{a+b-c}{2} = ab - \frac{(a+b)^2 - c^2}{4} = \frac{4ab - (a^2 + 2ab + b^2) + c^2}{4} = \frac{c^2 - (a-b)^2}{4} = \frac{c-(a-b)}{2} \cdot \frac{c+(a-b)}{2} = (s-a)(s-b)$.Now,$\frac{ab - r_1 r_2}{r_3} = \frac{(s-a)(s-b)}{\frac{\Delta}{s-c}} = \frac{(s-a)(s-b)(s-c)}{\Delta} = \frac{\Delta^2}{s \Delta} = \frac{\Delta}{s} = r$.

List-$I$	List-$II$
$(A)$ $rr_2 = r_1r_3$	$(I)$ $\angle A = 90^{\circ}$
$(B)$ $r_1 + r_2 = r_3 - r$	$(II)$ $b^2 = c^2 + a^2$
$(C)$ $r_1 = r + 2R$	$(III)$ $\angle C = 90^{\circ}$
	$(IV)$ $\angle B = 120^{\circ}$

In $\triangle ABC$,suppose the radius of the excircle opposite to angle $A$ is denoted by $r_1$,similarly $r_2$ for angle $B$,and $r_3$ for angle $C$. If $r$ is the radius of the inscribed circle,then what is the value of $\frac{ab - r_1 r_2}{r_3}$?

Similar Questions

In a $\triangle ABC$,which of the following formulae are correct?
$I. r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$II. r_1 = (s-a) \tan \frac{A}{2}$
$III. r_3 = \frac{\Delta}{s-c}$

In $\triangle ABC$, if $a : b : c = 4 : 5 : 6$, then the ratio of the circumradius to its inradius is

$AD, BE$ and $CF$ are the perpendiculars from the vertices of a $\Delta ABC$ to the opposite sides. The ratio of the perimeter of $\Delta DEF$ to the perimeter of $\Delta ABC$ is: (where $r$ is the inradius and $R$ is the circumradius of $\Delta ABC$)

In a $\triangle ABC$,which of the following formulae are correct?
$I. r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$II. r_1 = (s-a) \tan \frac{A}{2}$
$III. r_3 = \frac{\Delta}{s-c}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

In $\triangle ABC$,suppose the radius of the excircle opposite to angle $A$ is denoted by $r_1$,similarly $r_2$ for angle $B$,and $r_3$ for angle $C$. If $r$ is the radius of the inscribed circle,then what is the value of $\frac{ab - r_1 r_2}{r_3}$?

Similar Questions

In a $\triangle ABC$,which of the following formulae are correct?$I. r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$II. r_1 = (s-a) \tan \frac{A}{2}$$III. r_3 = \frac{\Delta}{s-c}$

In $\triangle ABC$, if $a : b : c = 4 : 5 : 6$, then the ratio of the circumradius to its inradius is

$AD, BE$ and $CF$ are the perpendiculars from the vertices of a $\Delta ABC$ to the opposite sides. The ratio of the perimeter of $\Delta DEF$ to the perimeter of $\Delta ABC$ is: (where $r$ is the inradius and $R$ is the circumradius of $\Delta ABC$)

In a $\triangle ABC$,which of the following formulae are correct?$I. r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$$II. r_1 = (s-a) \tan \frac{A}{2}$$III. r_3 = \frac{\Delta}{s-c}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

In a $\triangle ABC$,which of the following formulae are correct?
$I. r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$II. r_1 = (s-a) \tan \frac{A}{2}$
$III. r_3 = \frac{\Delta}{s-c}$

In a $\triangle ABC$,which of the following formulae are correct?
$I. r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$
$II. r_1 = (s-a) \tan \frac{A}{2}$
$III. r_3 = \frac{\Delta}{s-c}$