For a triangle with side lengths 6,5,and 9,fi | vedclass

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(B) Let the sides of the triangle be $a = 6$,$b = 5$,and $c = 9$. The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2} = \frac{6+5+9}{2} = \frac{20

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$\sqrt{2}$

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(B) Let the sides of the triangle be $a = 6$,$b = 5$,and $c = 9$. The semi-perimeter $s$ is given by $s = \frac{a+b+c}{2} = \frac{6+5+9}{2} = \frac{20}{2} = 10$. The area of the triangle $\Delta$ is calculated using Heron's formula: $\Delta = \sqrt{s(s-a)(s-b)(s-c)}$. $\Delta = \sqrt{10(10-6)(10-5)(10-9)} = \sqrt{10 	imes 4 	imes 5 	imes 1} = \sqrt{200} = 10\sqrt{2}$. The inradius $r$ is given by $r = \frac{\Delta}{s}$. $r = \frac{10\sqrt{2}}{10} = \sqrt{2}$.

For a triangle with side lengths $6$,$5$,and $9$,find the inradius of the triangle.

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