SOLUTION: (1.) The height of a triangle is 4 feet less than twice the base. If the area is 48 square feet, find the base and height. (2.) The height of an object projected vertically is g

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: (1.) The height of a triangle is 4 feet less than twice the base. If the area is 48 square feet, find the base and height. (2.) The height of an object projected vertically is g      Log On

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Question 447573: (1.) The height of a triangle is 4 feet less than twice the base. If the area is 48 square feet, find the base and height.
(2.) The height of an object projected vertically is given by h=24t-16t^2. At what times will the object be on the ground?
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
(1.) The height of a triangle is 4 feet less than twice the base. If the area is 48 square feet, find the base and height.
Since area of triangle = (1/2)*base*height
Let b = base
then
2b-4 = height
(1/2)b(2b-4) = 48
b(2b-4) = 96
2b^2-4b = 96
b^2-2b = 48
b^2-2b-48 = 0
(b-8)(b+6) = 0
b = {-6, 8}
throw out the negative solution leaving:
b = 8 feet (base)
.
height:
2b-4 = 2(8)-4 = 12 feet
.
(2.) The height of an object projected vertically is given by h=24t-16t^2. At what times will the object be on the ground?
set h to zero and solve for t:
0=24t-16t^2
0=3t-2t^2
0=t(3-2t)
t = {0, 3/2}
solution: 0 seconds and 1.5 seconds