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The side lengths of triangle ABC are 6, 8, and 10. What is the inradius of triangle ABC?

Nov 14, 2018

#1
+107405
+1

We can use a little logic, here

Let the sides  of the triangle be  AB, BC, AC

We have a 6-8-10  right triangle

The  area  =  (1/2)  [ Product of the legs ]  =  (1/2) [ 8 * 6 ]  = 24

Let the center of the incircle  = O

We can divide the right triangle into  three smaller triangles  =  AOB, BOC and AOC

The inradius will be tangent to each side of triangle ABC, so it will form an altitude to each of the three smaller triangles

So.....the area of ABC  = (1/2) (inradius) [ AB + BC + AC]

24   =  (1/2) (inradius) [ AB + BC + AC ]

24 / (1/2) =  (inradius)  [ perimeter ]

48  =  (inradius)  [ 6 + 8 + 10 ]

48  = (inradius) [ 24 ]

inradius  =  48 / 24   =   2 units

Nov 14, 2018
#2
+1

Here is another way:

Area =Base x Height / 2

Area =8 x 6/2 =24 units^2

Perimeter =6 + 8+ 10 =24 units

Semi-perimeter =24/2 = 12 units.

Area = Semi -perimeter x inradius

The inradius of a polygon can be found by $$A=rs$$, where $$A$$ is the area, $$r$$ is the inradius, and $$s$$ is the semi-perimeter. We know that the area is $$\frac{1}{2}*6*8=24$$, based on the information given, and $$\frac{6+8+10}{2}=12$$ which is considered the semi-perimeter. Now, we have to solve for the radius, and plugging our values in, we reach $$24=12r, 12r=24, \boxed{r=2}$$
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