The area of a right-angled triangle is 12 cm squared and one of the legs is 2 cm larger than the other.

The area of a right-angled triangle is 12 cm squared and one of the legs is 2 cm larger than the other. Find the length of the smaller leg. Let x be the length of the smaller leg.

Let’s denote the legs of this right-angled triangle through x and y.
According to the condition of the problem, one of the legs of this right-angled triangle is 2 cm larger than the other, therefore, the following relationship is true:
x = y + 2.
It is also known that the area of ​​this right-angled triangle is 12 square cm, therefore, the following relationship is true:
x * y / 2 = 12.
We solve the resulting system of equations. Substituting into the second equation the value x = y + 2 from the first equation, we get:
(y + 2) * y / 2 = 12.
We solve the resulting equation:
(y + 2) * y = 12 * 2;
y ^ 2 + 2 * y = 24;
y ^ 2 + 2 * y – 24 = 0.
The roots of this quadratic equation are guessed using Vieta’s theorem. Their sum must be -2, and the product must be -24. These numbers are 4 and -6.
Since leg length is positive, -6 is not appropriate. Hence:
y = 4.
Knowing y, we find x:
x = y + 2 = 4 + 2 = 6.

Answer: the legs of this right-angled triangle are 6 cm and 4 cm.