In an isosceles triangle, the side is 10 cm, and the base is 12. The circle centered on the side

In an isosceles triangle, the side is 10 cm, and the base is 12. The circle centered on the side of the triangle touches its other two sides. find the radius of the circle.

AC = CB = 10 cm;

AB = 12 cm.

Sides AC and BC are two tangents to the circle.

They are perpendicular to the radii DE and DF and form equal angles with the line CD through the center of the circle. This means that CD is the bisector of angle C. For an isosceles triangle ABC, it will also be the median:

AD = DB; AD + DB = AB; DB = AB / 2 = 12/2 = 6 cm.

CD = √ (CB ^ 2 – DB ^ 2) = √ (10 ^ 2 – 6 ^ 2) = 8 cm.

For right-angled triangles CBD and CED, the angle BCD is common. This means that these triangles are similar, and their sides are proportional:

CD / DE = CB / DB;

CD * DB = DE * CB;

DE = (CD * DB) / CB = (8 * 6) / 10 = 4.8 cm.

Answer: 4.8.