The hypotenuse of a right-angled triangle is 41cm, and its area is 180cm2. Find the legs of this triangle.

The area of ​​a right-angled triangle is half the product of the legs:

S = 0.5 * a * b;

a * b = 2 * S = 2 * 180 = 360.

The square of the hypotenuse is equal to the sum of the squares of the legs:

a ^ 2 + b ^ 2 = 41 ^ 2 = 1681.

We get the system of equations:

1) a * b = 360;

2) a ^ 2 + b ^ 2 = 1681.

We multiply both sides of the first equation by 2, add the result to the second equation:

2 * a * b + a ^ 2 + b ^ 2 = 720 + 1681;

(a + b) ^ 2 = 2401;

a + b = √2401 = 49.

Hence, a = 49 – b, we substitute the obtained value for a into the first equation of the system:

b * (49 – b) = 360;

49 * b – b ^ 2 = 360;

b ^ 2 – 49b +360 = 0.

We solve the quadratic equation:

D = 492 – 4 * 360 = 2401 – 1440 = 961 = 312.

b1 = (49 – 31) / 2 = 18/2 = 9;

b2 = (49 + 31) / 2 = 80/2 = 40.

With b = 9, a = 49 – 9 = 40.

For b = 40, a = 49 – 40 = 9.

Therefore, the legs of this triangle are 9 cm and 40 cm.