In a right-angled triangle ABC, the radius of the inscribed circle is 3 cm, the hypotenuse AB = 15 cm. Find AK and BK.

From the point O we construct the radii OH, OM to the points of tangency.

Quadrangle ОМСН square, CM = CM = R = 3 cm.

Let the length of the segment BK = X cm, then, by the property of a tangent drawn from one point, BH = BK = X cm, AK = AM = (15 – X) cm.

Then BC = (X + 3) cm, AC = (15 – X + 3) = (18 – X) cm.

By the Pythagorean theorem, AB ^ 2 = BC ^ 2 + AC ^ 2.

225 = (X + 3) ^ 2 + (18 – X) ^ 2.

225 = X ^ 2 + 6 * X + 9 + 324 – 36 * X + X ^ 2.

2 * X ^ 2 – 30 * X + 108 = 0.

X ^ 2 – 15 * X + 54 = 0.

Let’s solve the quadratic equation.

X1 = 6 cm.

X2 = 9 cm.

Answer: If AK = 6 cm, BC = 9 cm, if AK = 9 cm, BK = 6 cm.