The hypotenuse of a right triangle measures 10m. One leg of the triangle is 2m longer then the other. Find the lengths of the legs.
How would you solve this equation?
thank you :)
To solve this equation, you can use the Pythagorean theorem. The theorem states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the two legs.
Let's assume one leg of the triangle measures x meters. According to the problem, the other leg is 2 meters longer than x, so its length can be represented as (x + 2) meters.
Using the Pythagorean theorem, we have:
(x^2) + ((x + 2)^2) = (10^2)
Simplifying this equation, we get:
x^2 + (x^2 + 4x + 4) = 100
Combining like terms:
2x^2 + 4x + 4 = 100
Simplifying further:
2x^2 + 4x - 96 = 0
Now, we can solve this quadratic equation to find the values of x.
To solve this problem, we can start by assigning variables to the lengths of the legs of the right triangle. Let's call one leg "x" and the other leg "x+2" since the length of one leg is 2m longer than the other.
According to the Pythagorean theorem, which states that the square of the hypotenuse is equal to the sum of the squares of the other two sides, we can set up the following equation:
x^2 + (x+2)^2 = 10^2
Now we can simplify and solve for the lengths of the legs:
x^2 + (x^2 + 4x + 4) = 100
Combining like terms:
2x^2 + 4x + 4 = 100
Moving 100 to the left side of the equation:
2x^2 + 4x - 96 = 0
Dividing through by 2 to simplify:
x^2 + 2x - 48 = 0
Now we can factor the quadratic equation:
(x + 8)(x - 6) = 0
Setting each factor equal to zero:
x + 8 = 0 or x - 6 = 0
Solving for x:
x = -8 or x = 6
Since a length cannot be negative, we discard the negative solution. Therefore, the length of one leg is 6m, and the length of the other leg (which is 2m longer) is 8m.
Therefore, the lengths of the legs of the right triangle are 6m and 8m.
First of all, you have to write the equation. Call the shorter leg of the triangle x.
x^2 + (x+2)^2 = 10^2 = 100
2x^2 + 4x + 4 = 100
x^2 + 2x -48 = 0
(x+8)(x-6) = 0
Solve for x etc