A circle is inscribed in a right-angled triangle. The point of its contact with the hypotenuse divides

A circle is inscribed in a right-angled triangle. The point of its contact with the hypotenuse divides the hypotenuse into parts, the lengths of which are 6 cm and 4 cm. Find the area of this triangle.

Triangle ABC – rectangular, angle C = 90 degrees. A circle inscribed in ABC touches side AB at point K, side AC at point M, side BC at point P, AK = 6 cm, ВK = 4 cm.
The segments of the tangents to the circle drawn from one point are equal, therefore:
AM = AK = 6 cm;
ВK = BP = 4 cm;
CM = CP = x.
Since AB lies opposite the angle C, then AB is the hypotenuse.
By the Pythagorean theorem:
AB ^ 2 = AC ^ 2 + BC ^ 2.
AB consists of segments AK and ВK, then AB = 6 + 4 = 10 (cm).
AC consists of segments AM and CM, then: AC = AM + CM = 6 + x.
BC consists of segments ВР and СР, then: ВС = ВР + СР = 4 + x.
Substitute the known values ​​into the expression by the Pythagorean theorem and find the length x:
10 ^ 2 = (6 + x) ^ 2 + (4 + x) ^ 2;
36 + 12x + x ^ 2 + 16 + 8x + x ^ 2 = 100;
2x ^ 2 + 20x – 48 = 0;
x ^ 2 + 10x – 24 = 0.
Let’s solve the quadratic equation.
Discriminant:
D = b ^ 2 – 4ac;
D = 10 ^ 2 – 4 * 1 * (- 24) = 100 + 96 = 196.
x = (-b +/- √D) / 2a.
x1 = (-10 + √196) / 2 * 1 = (-10 + 14) / 2 = 4/2 = 2.
x2 = (-10 – √196) / 2 * 1 = (-10 – 14) / 2 = -24/2 = -12 – this value does not satisfy the meaning of the problem.
Hence:
CM = СР = 2 cm.
The area of ​​a right-angled triangle is equal to half the product of its legs:
S = (AC * BC) / 2;
S = (6 + 2) (4 + 2) / 2 = 8 * 6/2 = 48/2 = 24 (cm ^ 2).
Answer: S = 24 cm ^ 2.