Asked by jenn
the width of a basketball court is 1 m more than half the length. If the area of the court is 364 m squared, find the length and width. Sounds easy...but I do not know how to start off the problem.
Answers
Answered by
Jai
first we represent the unknowns using variables:
let x = length
according to first statement, the width is 1 more than half the length,, thus
let (1/2)x + 1 = width
now we set up the equation,, since we are only given the area, we can use this to solve for length and width. recall that the area of a rectangle (shape of the court) is given by
A = L*W
where L is length and W is width.
substituting,
A = L*W
364 = x*[ (1/2)x + 1 ]
364 = x*(0.5x + 1)
364 = 0.5 x^2 + x
we can transpose 364 to right side:
0 = 0.5x^2 + x - 364
we can multiply 2 to both sides of equation to make the 0.5 to whole number:
0 = x^2 + 2x - 728
we can either factor this or use quadratic formula,, but since it's factorable, we just factor this:
0 = (x-26)(x+28)
thus
x = 26 and x = -28
since measurements are always positive, we take only
x = 26 m (the length) ; and thus
(1/2)x + 1 = 14 m (the width)
hope this helps~ :)
let x = length
according to first statement, the width is 1 more than half the length,, thus
let (1/2)x + 1 = width
now we set up the equation,, since we are only given the area, we can use this to solve for length and width. recall that the area of a rectangle (shape of the court) is given by
A = L*W
where L is length and W is width.
substituting,
A = L*W
364 = x*[ (1/2)x + 1 ]
364 = x*(0.5x + 1)
364 = 0.5 x^2 + x
we can transpose 364 to right side:
0 = 0.5x^2 + x - 364
we can multiply 2 to both sides of equation to make the 0.5 to whole number:
0 = x^2 + 2x - 728
we can either factor this or use quadratic formula,, but since it's factorable, we just factor this:
0 = (x-26)(x+28)
thus
x = 26 and x = -28
since measurements are always positive, we take only
x = 26 m (the length) ; and thus
(1/2)x + 1 = 14 m (the width)
hope this helps~ :)
Answered by
:)
Thanks helped a lot
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