In a right-angled triangle, a circle is inscribed, the point of contact lying on the hypotenuse

In a right-angled triangle, a circle is inscribed, the point of contact lying on the hypotenuse divides it by 4cm and 6cm. Find the S (area) of the triangle.

Since the points H, K and M are the points of tangency of the triangle and the circle, then by the property of tangents drawn from one point: BK = BH = 6 cm, AM = AH = 4 cm, CK = CM.

OK and OM are the radii drawn to the points of tangency, then they are perpendicular to the tangents, and the quadrangle is OKCM square, with a side equal to the radius of the circle. CM = CK = R.

Then BC = (6 + R), AC = (4 + R).

Then, by the Pythagorean theorem: AB ^ 2 = BC ^ 2 + AC ^ 2.

100 = (6 + R) ^ 2 + (4 + R) ^ 2 = 36 + 12 * R + R ^ 2 + 16 + 8 * R + R ^ 2.

2 * R ^ 2 + 20 * R – 48 = 0.

R ^ 2 + 10 * R – 24 = 0

Let’s solve the quadratic equation.

R1 = -12 cm. (Not suitable as less than 0).

R2 = 2 cm.

Answer: The radius of the circle is 2 cm.