Find the dimensions of a rectangular piece of property that has a perimeter of 610 feet if the length is twenty feet more than two times the width

Question

Need Help on this word problem pls thank you 

Answer 1

Perimeter is defined as the sum of the 4 sides. A rectangle has lengths and widths that are equal. Thus the equation for perimeter of a rectangle is 2L+2W

 

2L+2W=610

 

L=2W+20 .. problem states the length is 20 feet more than 2 times the width

 

Plug that in for L

2(2W+20)+2W=610

4W+40+2W=610

6W+40=610

6W=570

W=90 feet

 

Plug that back into the equation 2L +2W=610

2L+180=610

2L=430

L=215 feet

 

Double check our answer: 

2(215) + 2(90)= 610

610=610

Answer 2

2L + 2W = 610

 

L = 2W + 20

 

2(2W + 20) + 2W = 610

 

4W +40 +2W = 610

 

6W = 570

 

W = 95

 

therefore L = 2*95 + 20 = 190 + 20 = 210

 

checking; 2*95 + 2*210 = 190 * 420 = 610 (proven)