The sides of a right angled triangle are equal to three consecutive nu...
To solve this problem, let's assume that the three consecutive numbers are x, x+1, and x+2.
According to the Pythagorean theorem, the sum of the squares of the two shorter sides of a right-angled triangle is equal to the square of the hypotenuse. In this case, we have:
x^2 + (x+1)^2 = (x+2)^2
Expanding and simplifying this equation, we get:
x^2 + x^2 + 2x + 1 = x^2 + 4x + 4
Combining like terms, we have:
2x^2 + 2x + 1 = x^2 + 4x + 4
Subtracting x^2 and 4x from both sides, we get:
x^2 - 2x - 3 = 0
Factoring this quadratic equation, we get:
(x - 3)(x + 1) = 0
So, x = 3 or x = -1. Since we are dealing with lengths, we can ignore the negative solution. Therefore, x = 3.
Thus, the sides of the right-angled triangle are 3 cm, 4 cm, and 5 cm.
The area of a right-angled triangle is given by the formula: Area = (1/2) * base * height.
In this case, the base and height of the right-angled triangle are 3 cm and 4 cm, respectively.
Therefore, the area of the right-angled triangle is:
Area = (1/2) * 3 cm * 4 cm = 6 cm^2
So, the area of such a triangle can be 6 cm^2, which corresponds to option (a).