SOLUTION: Perimeter of a high school basketball court. The perimeter of a standard high school basketball court is 268 ft. The length is 34 ft longer than the width. Find the dimensions of t

Click here to see ALL problems on Numeric Fractions

Question 443874: Perimeter of a high school basketball court. The perimeter of a standard high school basketball court is 268 ft. The length is 34 ft longer than the width. Find the dimensions of the court.
Source: Indiana High School Athletic Association
thank you!
Found 3 solutions by ewatrrr, chriswen, solver91311:
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website!

Hi

The perimeter of a standard high school basketball court is 268 ft.
The length is 34 ft longer than the width
Let x and (x+34) represent the width and length respectively
2*(x+34) + 2*x = 268ft
4x = 268 - 68
4x = 200
x = 50 ft, the width. the length is 84 ft
CHECKING our Answer***
2*84ft + 2*50ft = 268ft

ewatrrr@aol.com
private/pay tutoring Inquiries only, please.

Answer by chriswen(106)   (Show Source):
You can put this solution on YOUR website!
Let x ft be the width.
Let x+34 ft be the length.
P=2(l+w)
268=2(x+34+x)
268=2(2x+34) distribution law
268=4x+68
4x=268-68
4x=200
x=200/4
x=50
x+34=84
Therefore the width of the basketball court is 50ft and the length of the court is 84ft.

Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!

Let represent the width. Then must represent the length. Two times the width plus two times the length is the perimeter, so:

Solve for , then calculate .

John

My calculator said it, I believe it, that settles it