Find the legs of a right-angled triangle if their sum is 7 dm and the hypotenuse is 5 dm.

If the sum of two numbers is known, one of which we know, then in order to find the second number, it is necessary to subtract a known number from the sum of these numbers.

Let the first leg be equal to x dm, then the second leg is (7 – x) dm. Knowing the hypotenuse of the triangle, equal to 5 dm, we apply the Pythagorean theorem: The square of the hypotenuse is equal to the sum of the squares of the legs. The square of the hypotenuse is 5 ^ 2. The sum of the squares of the legs is (x ^ 2 + (7 – x) ^ 2). Let’s make an equation and solve it.

x ^ 2 + (7 – x) ^ 2 = 5 ^ 2;

x ^ 2 + 49 – 14x + x ^ 2 = 25;

2x ^ 2 – 14x + 49 – 25 = 0;

2x ^ 2 – 14x + 24 = 0;

x ^ 2 – 7x + 12 = 0;

D = b ^ 2 – 4ac;

D = (-7) ^ 2 – 4 * 1 * 12 = 49 – 48 = 1; √D = 1;

x = (-b ± √D) / (2a);

x1 = (7 + 1) / 2 = 8/2 = 4 (dm) – one first leg;

x2 = (7 – 1) / 2 = 6/2 = 3 (dm) – another first leg;

7 – x1 = 7 – 4 = 3 (dm) – one second leg;

7 – x2 = 7 – 3 = 4 (dm) – another second leg.

The legs can be 4 dm and 3 dm, or 3 dm and 4 dm, which is the same thing. 