If you have 14 feet of fencing, what are the areas of the different rectangles you could enclose with the fencing? Consider only whole number dimensions

Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²

Step-by-step explanation:

Feets of fencing = 14

Perimeter = 14

Let x = Length y = width

Perimeter = 2(x + y)

For x = 1

2(1 + y) = 14

1 + y = 7

y =6

Area of rectangle ; x *y = 1 * 6 = 6 ft²

For x = 2

2(2 + y) = 14

2 + y = 7

y = 5

Area of rectangle ; x *y = 2 * 5 = 10 ft²

For x = 3

2(3 + y) = 14

3 + y = 7

y =4

Area of rectangle ; x *y = 3 * 4 = 12 ft²

For x = 4

2(1 + y) = 14

4 + y = 7

y = 3

Area of rectangle ; x *y = 4 * 3 =12 ft²

Hence, Area of rectangle possible. : 6ft² ; 10 ft², 12 ft²

answer: c

step-by-step explanation: because you don't have to draw a point of intersection. couldyou give me brainliest?

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