Charles planted a garden using 36 feet of fencing for the perimeter. The area of Charles's garden is 45 square feet. What are the dimensions of his garden?
How Do You Find The Dimensions When You Know The Perimeter And The Square Feet?
Let P and A stand for the perimeter and area, respectively. Let L and W stand for the length and width, respectively.
We know ...
P = 2(L+W)
A = LW
The first equation can be solved for L, which can then be substituted into the second equation.
P/2 = L + W (divide both sides by 2)
P/2 - W = L (subtract W from both sides)
A = (P/2 - W)W (substitute for L)
A = WP/2 - W2 (use the distributive property)
W2 - WP/2 + A = 0 (add W^2 - WP/2 to both sides to put the quadratic into standard form)
W = (-(-P/2) ±√((-P/2)2 - 4(1)(A)))/(2(1))
= ((P/2) ±√((P2-16A)/4)/2
W = (P±√(P2-16A))/4
These two solutions turn out to be L and W.
Example
P=36 ft, A=45 ft2
W = (36±√(362-16*45))/4
= (36±√576)/4
= (36±24)/4
= {3, 15}
The width is 3 ft, the length is 15 ft.
We know ...
P = 2(L+W)
A = LW
The first equation can be solved for L, which can then be substituted into the second equation.
P/2 = L + W (divide both sides by 2)
P/2 - W = L (subtract W from both sides)
A = (P/2 - W)W (substitute for L)
A = WP/2 - W2 (use the distributive property)
W2 - WP/2 + A = 0 (add W^2 - WP/2 to both sides to put the quadratic into standard form)
W = (-(-P/2) ±√((-P/2)2 - 4(1)(A)))/(2(1))
= ((P/2) ±√((P2-16A)/4)/2
W = (P±√(P2-16A))/4
These two solutions turn out to be L and W.
Example
P=36 ft, A=45 ft2
W = (36±√(362-16*45))/4
= (36±√576)/4
= (36±24)/4
= {3, 15}
The width is 3 ft, the length is 15 ft.
