In a right triangle, the incircle touches the hypotenuse at a point that divides it into two segments measuring 5 cm and 12 cm. Determine the lengths of the legs of the triangle.
[In a right triangle, the point where the inscribed circle touches the hypotenuse divides it into segments of lengths 5 cm and 12 cm. Find the lengths of the triangle’s legs.]
Solution
According to the problem statement, AE = 5\,\text{cm}, BE = 12\,\text{cm}.
Let r be the radius of the incircle.
We have:
- BE = BP = 12\,\text{cm}
- PC = r
- BC = PC + BP = r + 12
- AE = AF = 5\,\text{cm}
- CF = r
- AC = CF + AF = r + 5
- AB = AE + EB = 5 + 12 = 17\,\text{cm}
In triangle \triangle ABC, angle \angle C = 90^\circ. By the Pythagorean theorem:
AB^2 = AC^2 + BC^2
(r + 12)^2 + (r + 5)^2 = 289
r^2 + 24r + 144 + r^2 + 10r + 25 = 289
2r^2 + 34r + 169 = 289
2r^2 + 34r - 120 = 0
r^2 + 17r - 60 = 0
Solving the quadratic equation gives:
r = 3\,\text{cm}
Then:
BC = r + 12 = 15\,\text{cm}, AC = r + 5 = 8\,\text{cm}
Answer:
8\,\text{cm};\quad 15\,\text{cm}
