Let's take a look at the given diagram. We are given that AB=AC=12 and BC=8, so the drawn triangle is . Additionally we know that all three segments are to ∘P.
First let's evaluate the height of the triangle h. To do this let's recall that the height in an isosceles triangle the base.
Let's write an equation for
h using the . Notice that since
h represents the height, we will consider only positive case when taking a of
h2.
42+h2=122
16+h2=144
h2=128
h=11.3137…
h≈11.3
The height of the triangle is approximately
11.3 units. Using this value, we can evaluate the .
A△ABC=21(11.3)(8)=45.2
The area of the triangle is
45.2 square units. Next let's recall that tangent segments are to the at the tangency points. Let's call the radius of the circle
r.
If we connect each vertex with the center of a triangle, then we have three triangles each with the height of r.
Notice that the sum of the areas of these three triangles needs to be equal to the area of
△ABC, 45.2. We will use the formula for the area of a triangle to write an equation and solve for
r.
A△ABC=A△BPC+A△BPA+A△APC⇓45.2=21(8)r+21(12)r+21(12)r
Let's solve the equation to find the radius.
45.2=21(8)r+21(12)r+21(12)r
45.2=28r+212r+212r
45.2=4r+6r+6r
45.2=16r
2.825=r
r=2.825
r≈2.8
The radius of the circle
P is approximately
2.8 units.