You can use the area method. First of all, find s that is half of perimeter:
[tex] s=\dfrac{5+7+6}{2} =9 [/tex] un.
Then:
1. Find the area of triangle using Heron's fofmula:
[tex] A=\sqrt{9\cdot (9-5)\cdot (9-7)\cdot (9-6)} =\sqrt{9\cdot4 \cdot 2\cdot 3} =6\sqrt{6} [/tex] sq. un.
2. Find the area of triangle using the formula [tex] A=s\cdot r [/tex], where r is the radius of inscribed circle.
[tex] A=9\cdot r [/tex] sq. un.
3. Equate both expressions:
[tex] 6\sqrt{6} =9r,\\ r=\dfrac{6\sqrt{6}}{9}=\dfrac{2}{3} \sqrt{6} [/tex] un.
Answer: [tex] r=\dfrac{2}{3} \sqrt{6} [/tex] un.
General Idea:
There are two formula's to find the area of triangle when three sides are given.
FIRST FORMULA: When a, b and c are the three sides of the triangle, we can use the Heron's triangle to find the Area of triangle.
[tex] Area \ of\ triangle =\sqrt{s(s-a)(s-b)(s-c)} ,\\\\ \ where\ s= \frac{a+b+c}{2} [/tex]
SECOND FORMULA: When the circle is inscribed in a triangle with radius r, then we can use the below formula to find the Area of triangle.
[tex] Area\ of\ triangle=\frac{1}{2} \cdot(a+b+c)\cdot r [/tex]
Applying the concept:
In our problem we are given the side lengths of triangle as 5, 6, & 7
Area of triangle using first formula:
[tex] s=\frac{5+6+7}{2}=\frac{18}{2}=9\\ \\ Area=\sqrt{9(9-5)(9-6)(9-7)}=\sqrt{9(4)(3)(2)} \\\\Area=6\sqrt{6}\ square\ units. [/tex]
Area of triangle using second formula:
[tex] Area=\frac{1}{2}\cdot (5+6+7)\cdot r\\ \\ Area=\frac{1}{2} \cdot(18)\cdot r\\ \\ Area =9r [/tex]
Conclusion:
We need to set up an equation based on the results that we get when finding the area of triangle using first and second formula, we get..
[tex] 9r=6\sqrt{6}\ \{Dividing \ 9\ on \ both\ sides\}\\ \\ \frac{9r}{9} =\frac{6\sqrt{6}}{9}
\ \{Simplifying \ the \ fraction \ on \ both\ sides\}\\ \\ r=\frac{2\sqrt{6}}{3} [/tex]
