The median drawn from the vertex of the right angle splits the ABC triangle into 2 triangles. The perimeter of which is 8 and 9, find the sides.

1. In a right-angled triangle, the median drawn from the right angle is half the hypotenuse. The median CM divides the hypotenuse AB into segments AM and BM, then:
CM = AM = BM = x.
By condition:
– perimeter △ AMC = AM + CM + AC = x + x + AC = 2 * x + AC = 8;
– perimeter △ BMC = BM + CM + BC = x + x + BC = 2 * x + BC = 9.
By the Pythagorean theorem:
AC = √ (AB² – BC²) = √ ((AM + BM) ² – BC²) = √ ((x + x) ² – BC²) = √ ((2 * x) ² – BC²) = √ (4 * x² – BC²).
2. Let’s solve a system of three equations:
2 * x + AC = 8;
2 * x + BC = 9;
AC = √ (4 * x² – BC²).
We substitute the AC expression from the third equation into the first:
2 * x + √ (4 * x² – BC²) = 8;
√ (4 * x² – BC²) = 8 – 2 * x;
4 * x² – BC² = 64 – 32 * x + 4 * x²;
– BC² = 64 – 32 * x;
BC² = 32 * x – 64;
BC = √ (32 * x – 64).
Substitute the resulting expression BC into the second equation:
2 * x + √ (32 * x – 64) = 9;
√ (32 * x – 64) = 9 – 2 * x;
32 * x – 64 = 81 – 36 * x + 4 * x²;
4 * x² – 36 * x – 32 * x + 81 + 64 = 0;
4 * x² – 68 * x + 145 = 0.
D = 68² – 4 * 4 * 145 = 4624 – 2320 = 2304.
x₁ = (68 – √2304) / (2 * 4) = (68 – 48) / 8 = 20/8 = 2.5.
x ₂ = (68 + √2304) / (2 * 4) = (68 + 48) / 8 = 116/8 = 14.5.
3. Find the lengths of the sides BC and AC.
3.1. 2 * x₁ + AC₁ = 8;
2 * 2.5 + AC₁ = 8;
AC₁ = 8 – 5;
AC₁ = 3.
2 * x₂ + AC₂ = 8;
2 * 14.5 + AC₂ = 8;
AC₂ = 8 – 29;
AC₂ = – 21 makes no sense.
3.2. 2 * x₁ + BC₁ = 9;
2 * 2.5 + BC₁ = 9;
BC₁ = 9 – 5;
BC₁ = 4.
2 * x₂ + BC₂ = 9;
2 * 14.5 + BC₂ = 9;
BC₂ = 9 – 29;
BC₂ = – 20 makes no sense.
4. By the Pythagorean theorem:
AB = √ (AC² + BC²) = √ (3² + 4²) = √ (9 + 16) = √25 = 5.
Answer: AC = 3, BC = 4, AB = 5.