Question

The length of a rectangular garden is 8 feet longer than it’s width. The garden is surrounded by a sidewalk that is 4 feet wide and has a area of 320 square feet. Find the dimensions of the garden.

Answer 1

Answer:

The length = 20
The width = 12 

Explanation:

Let the Length of the garden be L and the Width W 
Therefore the area of the garden = L*W 
But we know that L = W + 8 
Therefore the area of the garden can be expressed as W*(W + 8) 
When the brackets are expanded this equals W^2 + 8W

The area of the recctangle which includes the path and garden will have a length of L + 8 (ie the length of the garden + 4 feet at the top and 4 feet at the bottom) 
The width will be W + 8 (width of garden + 4 feet at the left and 4 feet at the right) 
Therefore the area will be (W + 8)*(L +8) 
Once again we know that L = W + 8 
Therefore the area of the path/garden = (W +8)(W +8 +8) 
=(W +8)(W +16) 
=W^2 +24W + 128 

We know that the path alone has an area of 320 square feet. Therefore if we subtract the area of the garden (W^2 + 8W) from the area of the path/garden the area left is the area of the path only 

Therefore W^2 + 24W + 128 - (W^2 + 8W) = 320 
W^2 + 24W + 128 - W^2 - 8W = 320 
Simplify 
16W + 128 = 320 
Subtract 128 from both sides of the equation 
16W = 192 
divide both sides of the equation by 16 
W = 12 
As L = W + 8 
L = 12 + 8 = 20