The hypotenuse of a right triangle is more than one leg by 9m and more than the other by 2m. Find the sides.

Let the hypotenuse of a triangle be equal to x m.Then its first leg is (x – 9) m, and its second leg – (x – 2) m.For any right-angled triangle, the Pythagorean theorem is true, according to which

(x – 9) ² + (x – 2) ² = x².

Let’s solve this equation and find the hypotenuse:

(x – 9) ² + (x – 2) ² = x²,

x² – 18x + 81 + x² – 4x + 4 – x²,

x² – 22x + 85 = 0,

D = (- 22) ² – 4 * 1 * 85 = 484 – 340 = 144,

x1,2 = (22 ± √144) / (2 * 1),

x1,2 = (22 ± 12) / 2,

x1 = (22 + 12) / 2 and x2 = (22 – 12) / 2,

x1 = 17 m and x2 = 5 m.

Let’s calculate what the legs are equal for the first option. Let’s find the first leg:

17 – 9 = 8 m.

Let’s find the second leg:

17 – 2 = 15 m.

Let’s calculate what the legs are equal in the second option. Let’s find the first leg:

5 – 9 = – 4 m.

Length cannot be negative. This means that the second option is, in principle, not correct and it makes no sense to consider it further.

Answer: in a right-angled triangle, the hypotenuse is 17 m, the first leg is 8 m and the second leg is 15 m.