The hypotenuse of a right triangle is 32 cm larger than one of the legs and 9 cm larger than the other

The hypotenuse of a right triangle is 32 cm larger than one of the legs and 9 cm larger than the other. Find the sides of the triangle.

Let’s denote by x the length of the hypotenuse of this right-angled triangle.

According to the condition of the problem, the hypotenuse of this right-angled triangle is larger than one of the legs by 32 cm and more than the other leg by 9 cm, therefore, the lengths of the legs of this right-angled triangle are x – 32 cm and x – 9 cm, and using the Pythagorean theorem, we obtain the following equation:

(x – 32) ^ 2 + (x – 9) ^ 2 = x ^ 2.

We solve the resulting equation:

x ^ 2 – 64x + 1024 + x ^ 2 – 18x + 81 = x ^ 2;

2x ^ 2 – x ^ 2 – 82x + 1105 = 0;

x ^ 2 – 82x + 1105 = 0;

x = 41 ± √ (1681 – 1105) = 41 ± √576 = 41 ± 24.

x1 = 41 – 24 = 17;

x2 = 41 + 24 = 65.

We find the legs at x = 17:

x – 32 = 17 – 32 = -15.

Since the length of the leg is positive, the value x = 17 is not suitable.

We find the legs at x = 65:

x – 32 = 65 – 32 = 33;

x – 9 = 65 – 9 = 56.

Answer: the legs of this right-angled triangle are 33 cm and 56 cm, the hypotenuse of this right-angled triangle is 65 cm.