Anyone good at algebra?
Because I know I'm not. Having a hard time with this one, which is probably cake to the rest of you. help me out?
One leg of a right triangle is 2 mm shorter than the longer leg and the hypotenuse is 2 mm longer than the longer leg. What are the lengths of the triangle?
can you solve it?
December 11, 2014
5 Comments • Newest first
You could try and guess the lengths.
Should be a 6-8-10 triangle.
So basically:
x x-2
x+2
(x)^2 + (x-2)^2 = (x+2)^2 Due to pythagorean theorem.
X^2 + x^2 -4x + 4 = x^2 + 4x + 4
Now just simplify.
Edit: What class are you taking in High school right now ?
Just curious.
Reference leg: x
Shorter leg: (x-2)
Hypotenuse: (x+2)
You can then use the Pythagorean Theorem to solve it
(x)^2 + (x-2)^2 = (x+2)^2
Apply FOIL, move all variables to one side and numbers to the other to solve for x. When you have x, plug x into the respective legs I posted earlier to find the lengths of all the legs. Remember all of this is done in mm, never forget your variables
(x-2)^2 x (x)^2) = (x+2)^2
I think this is how you solve it. Expand on the equation, then determine x. You can substitute x afterwards to find the lengths of the triangle
Short leg = x-2
long leg = x
hypotenuse = x+2