The area of a right-angled triangle is 24 ^ 2 and its hypotenuse is 10cm what are the legs of the triangle.

Let us denote (in centimeters) through a, b – legs and through c – the hypotenuse of this right-angled triangle.
As you know, the area of ​​a right-angled triangle is equal to half the product of the legs of the triangle, that is, (½) * a * b = 24 or a * b = 48.
According to the Pythagorean theorem, a² + b² = c². By the condition of the task, the hypotenuse of the triangle is 10 cm. Hence, a² + b² = 10² or a² + b² = 100.
Thus, we got the following system of equations of the second degree: a * b = 48 and a² + b² = 100. Let’s solve this system. Multiply the first equation by 2 and find the algebraic sum of the resulting and the second equation: a² + 2 * a * b + b² = 100 + 2 * 48 or using the abbreviated multiplication formula (a + b) 2 = a2 + 2 * a * b + b2 (the square of the sum), (a + b) 2 = 14². Taking the arithmetic square root on both sides of the last equality, we get a + b = 14.
The equalities a + b = 14 and a * b = 48, according to Vieta’s theorem, allow us to assert that a and b are the roots of the quadratic equation x² – 14 * x + 48 = 0. Let’s solve it. For this purpose, we calculate the discriminant D = (–14) ² – 4 * 1 * 48 = 196 – 192 = 4> 0. This means that the quadratic equation has two real roots: х1 = (14 – √ (4)) / 2 = 6 and x2 = (14 + √ (4)) / 2 = 8.
Thus, the legs of this triangle are 6 cm and 8 cm.
Answer: 6 cm and 8 cm.