Let's label the center of the circle O. We can see that triangle ABC is a right triangle with angle B being the right angle.
First, let's find the area of triangle ABC using Heron's formula:
s = (AB + BC + CA) / 2
s = (6 + 10 + 16) / 2
s = 16
Area of triangle ABC = sqrt (s(s-AB)(s-AC)(s-BC))
Area of triangle ABC = sqrt (16(16-6)(16-16)(16-10))
Area of triangle ABC = sqrt (16 * 10 * 6 * 6)
Area of triangle ABC = sqrt (5760)
Area of triangle ABC = 24
Since the radius of the circle is the same as the height of triangle ABC drawn from point O, and the area of a triangle is equal to 0.5 * base * height, we can find the height by dividing the area by the base:
Height = 2 * Area / AB
Height = 2 * 24 / 6
Height = 8
Now, we have a right triangle OBC with OB as the radius of the circle, BC as the base, and height as 8. Using the Pythagorean theorem:
OB^2 = BC^2 + Height^2
OB^2 = 10^2 + 8^2
OB^2 = 100 + 64
OB^2 = 164
OB ≈ sqrt(164)
OB ≈ 12.81
Therefore, the radius of the circle is approximately 12.81.
Points A , B, and C are drawn, with the following:
AB = 6
AC = 16
BC = 10
Point B is on AC
Semicircles are constructed on AB, AC, and BC, and another circle is constructed such that it is tangent to all three semicircles. Find the radius of the circle.
1 answer
1 answer