The hypotenuse of a right-angled triangle is 9 cm larger than one of its legs and 32 cm

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

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

In the condition of the problem it is said that the hypotenuse of this right-angled triangle is more than one of its legs by 9 cm and more than its other leg by 32 cm, therefore, the legs of this triangle are x – 9 cm and x – 32 cm, and using the Pythagorean theorem, we obtain the following equation :

(x – 9) ^ 2 + (x – 32) ^ 2 = x ^ 2.

We solve the resulting equation:

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

x ^ 2 – 82x + 1105 = 0;

x = 41 ± √ (41 ^ 2 – 1105) = 41 ± √ (1681 – 1105) = 41 ± √576 = 41 ± 24.

x1 = 41 + 24 = 65;

x2 = 41 – 24 = 17.

Since the hypotenuse is 32 cm larger than one of the legs, the value x = 17 is not suitable.

Thus, the length of the hypotenuse of this right-angled triangle is 65 cm.

We find the lengths of the legs:

x – 9 = 65 – 9 = 56 cm;

x – 32 = 65 – 32 = 33 cm.

Answer: the lengths of the sides of this triangle are 33 cm, 56 cm and 65 cm.