The sides of a right angled triangle are x,x+1 and x+2.find x,Am not gettin the right answer.
Being a right triangle, x^2 + (x + 1)^2 = (x + 2)^2.
Expanding and simplifying, yields x^2 -- 2x - 3 = 0.
Factoring or the quadratic equation will lead you to the smallest Pythagorean Triple triangle with sides of 3, 4 and 5.
To solve this problem, we need to apply the Pythagorean theorem, which states that in a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse.
In this case, the sides of the right triangle are given as x, x+1, and x+2. Let's use these values to set up the equation:
(x)^2 + (x+1)^2 = (x+2)^2
Now, let's simplify the equation step by step:
x^2 + (x+1)(x+1) = (x+2)(x+2)
x^2 + (x^2 + 2x + 1) = (x^2 + 4x + 4)
x^2 + x^2 + 2x + 1 = x^2 + 4x + 4
2x^2 + 2x + 1 = x^2 + 4x + 4
2x^2 + 2x + 1 - x^2 - 4x - 4 = 0
x^2 - 2x - 3 = 0
Now we have a quadratic equation in terms of x. We can solve this equation using factoring or the quadratic formula.
Factoring:
(x - 3)(x + 1) = 0
Setting each factor equal to zero, we have:
x - 3 = 0 or x + 1 = 0
Solving each equation, we find:
x = 3 or x = -1
Since the side lengths of a triangle cannot be negative, we discard the solution x = -1.
Therefore, the value of x is 3.
Plug this value back into the original expression to check:
x = 3
x+1 = 4
x+2 = 5
So, the sides of the right triangle are 3, 4, and 5.