A rectangular field is twelve times as long as it is wide. If the perimeter of the field is 1300 feet, what are the dimensions of the field?

Question

A rectangular field is twelve times as long as it is wide. If the perimeter of the field is 1300 feet, what are the dimensions of the field?

Answer 1

Perimiter of a Rectangle is 2*(L+W)

If L is the length , W is the width of the Rectangle

so L is 12 times of its widith

L= 12W as from the condition given,

Perimiter is 1300 meter given

Therfore we can write the equation from the formula of perimeter 2*(12W+W) = 1300

2* 13W= 1300

13W= 1300/2

13W= 650

dividing both sides by 13 we get,

W= 650/13=50 , therefore L=12*50=600 hence, Length is 600 ft and wide is 50 ft dimension of the field is 600ft by 50 ft

Answer 2

Lets say the length of the field is "x" and the width is "y".

We can make the two following equations using the information in the question:

x = 12y

2x + 2y = 1300

Since the first equation gives us a value for x (= 12y), we can substitute that in for x in the second equation to get:

2(12y) + 2y = 1300 Simplified to:

24y+ 2y = 1300

26y = 1300

Divide both sides by 26 to get:

y = 50 feet

We can then use this in the first equation to get

x = 12(50)

x = 600

So the field is 600 feet by 50 feet.