In triangle ABC,if a=6, b=8 and c=10,then fra | vedclass

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

Question

(A) Given $a=6, b=8, c=10$. Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$,the triangle is a right-angled triangle with the hypotenuse $c=10$.The semi-perim

Vedclass · Accepted Answer

$b+c$

Vedclass · Answer

(A) Given $a=6, b=8, c=10$. Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$,the triangle is a right-angled triangle with the hypotenuse $c=10$.The semi-perimeter $s = \frac{a+b+c}{2} = \frac{6+8+10}{2} = 12$.The area of the triangle $\Delta = \frac{1}{2} 	imes 6 	imes 8 = 24$.The exradii are given by $r_1 = \frac{\Delta}{s-a} = \frac{24}{12-6} = 4$,$r_2 = \frac{\Delta}{s-b} = \frac{24}{12-8} = 6$,$r_3 = \frac{\Delta}{s-c} = \frac{24}{12-10} = 12$.The inradius $r = \frac{\Delta}{s} = \frac{24}{12} = 2$.Now,calculate the expression $\frac{2 r_2 r_3}{r r_1} = \frac{2 	imes 6 	imes 12}{2 	imes 4} = \frac{144}{8} = 18$.Check the options: $a+b = 6+8 = 14$,$b+c = 8+10 = 18$,$c+a = 10+6 = 16$,$a+b+c = 24$.Thus,the value is $b+c = 18$.

In triangle $ABC$,if $a=6, b=8$ and $c=10$,then $\frac{2 r_2 r_3}{r r_1} = $

Similar Questions

In a triangle $ABC$, $a=5$, $b=4$ and $\tan \frac{C}{2}=\sqrt{\frac{7}{9}}$, then its inradius $r=$

$AD, BE$ 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 $\triangle ABC$, find the value of $r_1+r_2+r_3$.

If in a $\triangle ABC$, $AD$, $BE$, and $CF$ are the altitudes and $R$ is the circumradius of $\triangle ABC$, then the radius of the circumcircle of $\triangle DEF$ is

If $R$ is the radius of the circumcircle of the $\Delta ABC$ and $\Delta$ is its area,then

Vedclass Test Series

Exam Paper Generator

Online Exam Module