Find the perimeter of the triangle if its hypotenuse is 25 and one of the legs is 17 larger than the other.

1. Vertices of a right-angled triangle. A, B, C. AB = 25 units. The length of the AC leg is 17 units longer than the BC leg.

2. We take the length of the BC leg as x (units). The length of the AC leg (x + 17) units.

3. AB² = AC² + BC².

4. We make the equation:

x² + (x + 17) ² = 25²;

2x² + 34x – 336 = 0;

x² + 17x – 168 = 0;

The first value x = (- 17 + √289 + 4 x 168) / 2 = (- 17 + 31) / 2 = 7 units.

The second value is x = (- 17 – 31) / 2 = – 24. Not accepted.

5. ВС = 7 units of measurement. AC = 7 + 17 = 24 units.

6. The perimeter of the triangle ABC = 7 + 24 + 25 = 56 units.

Answer: the perimeter of the triangle ABC = 56 units.