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

beastrohit Nov 14, 2018

#1**+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

CPhill 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

Inradius =Area / Semi-perimeter

Inradius =24 / 12 =2 units.

Guest Nov 14, 2018

#3**+1 **

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 \(24=12r, 12r=24, \boxed{r=2}\).

neworleans06 Nov 14, 2018