In a right-angled triangle, one leg is larger than the other by 7, and the hypotenuse

In a right-angled triangle, one leg is larger than the other by 7, and the hypotenuse is larger than the smaller leg by 8. Find the area of this triangle.

Let the smaller leg a = X, the larger b = X + 7, and the hypotenuse c = X + 8.
We apply the Pythagorean theorem for the hypotenuse, we have:
(X + 8) ^ 2 = (X + 7) ^ 2 + X ^ 2;
We apply the abbreviated multiplication formulas, the square of the binomial, we have:
X ^ 2 + 16X + 64 = X ^ 2 + 14X + 49 + X ^ 2;
We transfer all the terms of the equation from the right to the left by changing the signs to opposite ones and reduce similar terms:
X ^ 2 + 16X + 64-2X ^ 2-14X-49 = 0;
-X ^ 2 + 2X + 15 = 0; quadratic equation, multiply both sides by -1, we have:
X ^ 2-2X-15 = 0;
We apply Vieta’s Theorem to find the roots of a quadratic equation.
X1 = -3; X2 = 5.
We discard the root -3 negative, since the length cannot be negative.
a = 5;
b = 5 + 7 = 12;
c = 5 + 8 = 13.
Area formula S = 1/2 a * b = ½ * 5 * 12 = 30.