+159 The side lengths of triangle ABC are 6, 8, and 10. What is the inradius of triangle ABC?
+129852 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
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.
+54 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 \(24=12r, 12r=24, \boxed{r=2}\).
