The lengths ofthe sides of a right-angled triangle at (3x+ 1)cm, (3x - 1)cm and xcm.
and the question is???
Guessing that the question is to find the sides.
x^2 + (3x-1)^2 = (3x+1)^2
x^2 + 9x^2 - 6x + 1 = 9x^2 + 6x + 1
x^2 - 12x = 0
Use your favourite method to solve this quadratic,
remember that x > 0
(3x+ 1)^2 = (3x - 1)^2 + x^2
9x^2 + 6x + 1 = 9x^2 - 6x + 1 + x^2
x^2 - 12x = 0 ... x = 12
To find the lengths of the sides of a right-angled triangle, we can use the Pythagorean theorem, which states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides.
Let's denote the sides of the triangle as follows:
Side 1: (3x + 1) cm
Side 2: (3x - 1) cm
Hypotenuse: x cm
According to the Pythagorean theorem, we can write the equation as:
Hypotenuse^2 = Side 1^2 + Side 2^2
Substituting in the given values, we get:
x^2 = (3x + 1)^2 + (3x - 1)^2
Expanding the equation:
x^2 = (9x^2 + 6x + 1) + (9x^2 - 6x + 1)
Now, let's simplify the equation:
x^2 = 9x^2 + 6x + 1 + 9x^2 - 6x + 1
Combining like terms:
x^2 = 18x^2 + 2
Subtracting x^2 from both sides:
0 = 17x^2 + 2
This is a quadratic equation, but we need to verify if it has real solutions. To do this, we can calculate the discriminant (b^2 - 4ac) of the quadratic equation.
In this case, a = 17, b = 0, and c = 2.
Discriminant = 0^2 - 4(17)(2) = -136
Since the discriminant is negative, it means that the quadratic equation has no real solutions. Therefore, there are no values of x that satisfy the given lengths for the sides of a right-angled triangle.