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’s denote by x the smaller leg of this right-angled triangle.
According to the condition of the problem, in this right-angled triangle, one leg is greater than the other by 7, therefore, the second leg is x + 7.
It is also known that the hypotenuse in this triangle is 8 larger than the smaller leg, therefore, the hypotenuse is x + 8.
Using the Pythagorean theorem, we can draw up the following relationship:
x² + (x + 7) ² = (x + 8) ².
We solve the resulting equation:
x² = (x + 8) ² – (x + 7) ²;
x² = (x + 8 – x – 7) * (x + 8 + x + 7);
x² = 1 * (2x + 15);
x² = 2x + 15;
x² – 2x – 15 = 0;
x = 1 ± √ (1 + 15) = 1 ± √16 = 1 ± 4;
x1 = 1 + 4 = 5:
x2 = 1 – 4 = -3.
Since the length of the leg is positive, the value x = -3 is not suitable.
Knowing the smaller leg of a given right-angled triangle, we find the larger leg:
x + 7 = x + 5 = 12.
Find the area S of a given right-angled triangle as half the product of its legs:
S = 5 * 12/2 = 5 * 6 = 30.

Answer: The area of ​​this right-angled triangle is 30.