The area of a right triangle is 210 ^ 2 and its hypotenuse is 37 cm. Find the perimeter of this triangle.

Let the legs be equal to a and b. The area of ​​a right-angled triangle is 1/2 * a * b.

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

The system of equations will turn out:

1/2 * a * b = 210; a ^ 2 + b ^ 2 = 37 ^ 2.

a * b = 420; a ^ 2 + b ^ 2 = 1369.

We express a from the first equation and substitute it into the second:

a = 420 / b.

(420 / b) ^ 2 + b ^ 2 = 1369;

176400 / b ^ 2 + b ^ 2 = 1369; multiply by b ^ 2:

176400 + b ^ 4 = 1369b ^ 2;

b ^ 4 – 1369b ^ 2 + 176400 = 0.

Let’s introduce a new variable: let b ^ 2 = c.

c ^ 2 – 1369c + 176400 = 0.

D = 1874161 – 705600 = 1168561 (√D = 1081);

c1 = (-1369-1081) / 2 = 2450/2 = 1225.

c2 = (-1369 + 1081) / 2 = 288/2 = 144.

Let’s go back to replacing b ^ 2 = c:

b ^ 2 = 1225; b = 35.

b ^ 2 = 144; b = 12.

Let’s find the value of a: a = 420 / b.

b = 35, a = 420/35 = 12.

b = 12; a = 420/12 = 35.

The legs of a right-angled triangle are 12 and 35, and the hypotenuse is 37 (by condition).

We calculate the perimeter of the triangle: P = 12 + 35 + 37 = 84 cm.