The side lengths of triangle ABC are 6,8, and 10. What is the inradius of triangle ABC?
+121
The triangle ABC is made up of three smaller triangle, namely AIC, AIB, and BIC. These triangles have the same height, namely r, the inradius of the triangle, and their bases are \(a, b\ and\ c.\) Adding the areas of these triangle we get the area of the triangle ABC, and in terms of r, that area is
\(\frac{1}{2}ar+\frac{1}{2}br+\frac{1}{2}cr=\frac{a+b+c}{2} \cdot r\) \(=\frac{6+8+10}{2} \cdot r=12r\).
So if we can find the area of triangle ABC , then by dividing that area by 12, we would have the inradius. Fortunately Heron, or Hero of Alexandria, some Greek dude who lived around 10 AD, left us a formula by which we can calculate thee area we need using only the known sides of the triangle ABC. If we let \(s=\frac{a+b+c}{2}\), a quantity we calculated to be 12, the the area of ABC would be\(\sqrt {s(s-a)(s-b)(s-c)}\)\(=\sqrt{12(12-10)(12-8)(12-6)} = \sqrt{12\cdot48} =\sqrt{36\cdot16}=24\). So the measure of the inradius is 24 divided by 12 or 2 units of length.
+121
The triangle ABC is made up of three smaller triangle, namely AIC, AIB, and BIC. These triangles have the same height, namely r, the inradius of the triangle, and their bases are \(a, b\ and\ c.\) Adding the areas of these triangle we get the area of the triangle ABC, and in terms of r, that area is
\(\frac{1}{2}ar+\frac{1}{2}br+\frac{1}{2}cr=\frac{a+b+c}{2} \cdot r\) \(=\frac{6+8+10}{2} \cdot r=12r\).
So if we can find the area of triangle ABC , then by dividing that area by 12, we would have the inradius. Fortunately Heron, or Hero of Alexandria, some Greek dude who lived around 10 AD, left us a formula by which we can calculate thee area we need using only the known sides of the triangle ABC. If we let \(s=\frac{a+b+c}{2}\), a quantity we calculated to be 12, the the area of ABC would be\(\sqrt {s(s-a)(s-b)(s-c)}\)\(=\sqrt{12(12-10)(12-8)(12-6)} = \sqrt{12\cdot48} =\sqrt{36\cdot16}=24\). So the measure of the inradius is 24 divided by 12 or 2 units of length.
+2863 This is a formula that he uses \(A=rs\)
s = semi-perimeter
A = area
r = inradius
Using his method (heron's formula) we can find the value of A.
Using the given info on the triangles side, we can find the semi-perimeter by finding the perimeter and dividing by 2.
Then with that info, we plug it into the formula and solve for R.
