If the sides of the triangle are 5K, 6K, 5K a | vedclass

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

Question

(A) Let the sides of the triangle be $a = 5K$,$b = 6K$,and $c = 5K$.The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \frac{5K + 6K + 5K}{

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Let the sides of the triangle be $a = 5K$,$b = 6K$,and $c = 5K$.The semi-perimeter $s$ is given by $s = \frac{a + b + c}{2} = \frac{5K + 6K + 5K}{2} = \frac{16K}{2} = 8K$.The area of the triangle $\Delta$ using Heron's formula is $\Delta = \sqrt{s(s - a)(s - b)(s - c)}$.Substituting the values: $\Delta = \sqrt{8K(8K - 5K)(8K - 6K)(8K - 5K)} = \sqrt{8K 	imes 3K 	imes 2K 	imes 3K} = \sqrt{144K^4} = 12K^2$.The radius of the incircle $r$ is given by $r = \frac{\Delta}{s}$.Given $r = 6$,we have $6 = \frac{12K^2}{8K}$.Simplifying the expression: $6 = \frac{3K}{2}$.Therefore,$3K = 12$,which gives $K = 4$.

If the sides of the triangle are $5K, 6K, 5K$ and the radius of the incircle is $6$,then the value of $K$ is equal to

Similar Questions

If the angular bisector of the angle $A$ of the triangle $ABC$ meets its circumcircle at $E$ and the opposite side $BC$ at $D$, then $DE \cos \frac{A}{2} = $

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

In a $\triangle ABC$,if $a=13, b=14, c=15$,then find the value of $r_1$.

The sides of a right-angled triangle are integers. The length of one of the sides is $12$. The largest possible radius of the incircle of such a triangle is

If the sides of a triangle are $3$, $4$, and $5$, then the circumradius of the triangle is

Vedclass Test Series

Exam Paper Generator

Online Exam Module