SOLUTION: The length of a lacrosse field is 15 yards less than twice its width, and the perimeter is 330 yards. The defensive area of the field is 5/21 of the total field area. Find the defe

Algebra ->  Rectangles -> SOLUTION: The length of a lacrosse field is 15 yards less than twice its width, and the perimeter is 330 yards. The defensive area of the field is 5/21 of the total field area. Find the defe      Log On


   

Question 911226: The length of a lacrosse field is 15 yards less than twice its width, and the perimeter is 330 yards. The defensive area of the field is 5/21 of the total field area. Find the defensive area of the lacrosse field.
Found 2 solutions by ewatrrr, ichigo449:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

     .

L + w = 165

(2w-15) + w = 165

3w = 180

w = 60yd  and L = 105yd

A = 60yd%2A105yd+=+6300yd%5E2

%285%2F21%296300yd%5E2 = 1500yd^2

Answer by ichigo449(30) About Me  (Show Source):
You can put this solution on YOUR website!
Let x be the length of the field and y the width. The first condition in the question says that x= 2y-15. Then the perimeter gives us a linear equation:
2y + 2(2y-15) = 330
This can be simplified to 6y-30 = 330 or 6y = 300 so y = 60. Then, by x = 2y-15, x = 105. Now from area of a rectangle, x*y, we get the total field area to be: 6300. Since the defensive area is 5/21 the total area we conclude that it must be 1500 yards square.