SOLUTION: A public vegetable garden is divided up in different shapes for those who wish to grow vegetables there. Mitch’s piece is in the shape of a right triangle and has an area of 40 s

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: A public vegetable garden is divided up in different shapes for those who wish to grow vegetables there. Mitch’s piece is in the shape of a right triangle and has an area of 40 s      Log On


   

Question 1152370: A public vegetable garden is divided up in different shapes for those who wish to grow vegetables there. Mitch’s piece is in the shape of a right triangle and has an area of 40 square feet. The longer leg is 6 feet less than twice the shorter leg. What are the lengths of the two legs of that triangle?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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A public vegetable garden is divided up in different shapes for those who wish to grow vegetables there.
Mitch’s piece is in the shape of a right triangle and has an area of 40 square feet.
The longer leg is 6 feet less than twice the shorter leg.
What are the lengths of the two legs of that triangle?
:
let x = the length of the shorter leg (let this be the base)
then
(2x-6) = the length of the longer leg (let this be the height)
:
the area
1%2F2*b*h = 40
b*h = 80
Replace b and h
x(2x-6) = 80
2x^2 - 6x - 80 = 0
simplify, divide by 2
x^2 - 3x - 40 = 0
Factors to
(x-8)(x+5) = 0
the positive solution is what we want here
x = 8 ft is the shorter leg
then
2(8) - 6 = 10 ft is the longer leg
:
:
Check in the area equation
1%2F2*8*10 = 40