find the length of shorter leg of a right triangle if the longer leg is 20feet more than the shorter leg and the hypotenuse is 20feet less than twice the shorter. please explain this whole thing to help me solve it.
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it helpes to set the shorter leg to x.
so you get that the longer leg L = x + 20
and you get that the hypotenuse H = 2x - 20
now use that it is a right triangle so that
x^2 + L^2 = H^2 and fill in H and L with the x's
then solve for x and get the length of the shorter leg
The problem provides you with three equations. Let's say that the shorter leg is called A, the longer leg is called B, and the hypotenuse is C. Because it's a right triangle, we know that A^2 + B^2 = C^2. We know that the longer leg is 20 feet more than the shorter leg, so B = A + 20. We know that the hypotenuse is 20 feet less than twice the shorter leg, so C = 2A - 20.
Writing all these out gives us a system of equations:
A^2 + B^2 = C^2
B = A + 20
C = 2A - 20.
The second two equations give use expressions for B and C in terms of A. We can substitute these into the first equation to reduce it to an equation of only a single variable, A:
A^2 + (A+20)^2 = (2A-20)^2
A^2 + A^2 + 40A + 400 = 4A^2 - 80A + 400
2A^2 + 40A = 4A^2 - 80A.
120A = 2A^2.
A^2 = 60A.
A = 60.
Now we can substitute this value into our other two equations:
B = A + 20 = 60 + 20 = 80
C = 2A - 20 = 2* 60 - 20 = 100.
(A,B,C) = (60,80,100).
@stoopkid Don't completely solve the question. I thinkt that this person is genuine but don't give the full answer away ;)