The sides of a right angle triangle are xcm, (x+2)cm and (x+3)cm. Find the length of the longest and the shortest side
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First, find x.
x^2 + (x+2)^2 = (x+3)^2
x = 1+√6
But what is the figure for the largest
and which is for the shortest
To find the lengths of the longest and shortest sides of a right angle triangle with sides x cm, (x+2) cm, and (x+3) cm, we need to apply the Pythagorean theorem.
According to the Pythagorean theorem, in a right angle triangle, the square of the length of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides.
In this case, let's assume that (x+3) cm is the hypotenuse. Using the theorem, we can write the equation:
(x+3)^2 = x^2 + (x+2)^2
Expanding and simplifying the equation:
x^2 + 6x + 9 = x^2 + x^2 + 4x + 4
Combining like terms:
2x^2 + 10x + 9 = 2x^2 + 4x + 4
Simplifying further:
6x + 9 = 4x + 4
Bringing like terms to one side:
6x - 4x = 4 - 9
2x = -5
Dividing both sides by 2:
x = -5/2
Since a negative value is not meaningful for a length, this solution is not valid.
Therefore, there is no valid solution using (x+3) cm as the hypotenuse. However, we can try assuming (x+2) cm as the hypotenuse and (x+3) cm as the longest side.
Applying the Pythagorean theorem again, we have:
(x+2)^2 = x^2 + (x+3)^2
Expanding and simplifying:
x^2 + 4x + 4 = x^2 + x^2 + 6x + 9
Combining like terms:
4x + 4 = 2x^2 + 6x + 9
Simplifying further:
4x + 4 = 2x^2 + 6x + 9
Subtracting 4x and 4 from both sides:
0 = 2x^2 + 2x + 5
Now, we can use the quadratic formula to solve for x:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 2, b = 2, and c = 5.
Substituting these values into the quadratic formula:
x = (-(2) ± √((2)^2 - 4(2)(5))) / (2(2))
x = (-2 ± √(4 - 40)) / (4)
x = (-2 ± √(-36)) / 4
Since we have the square root of a negative number, there are no real solutions to this equation. Therefore, we conclude that there are no valid lengths for the sides of the right angle triangle with the given conditions.