Question

What is the length and width of a rectangle with a perimeter of 28 feet and an area of 48 sq feet

Answer 1

The rectangle is 8 ft by 6 ft.

Explanation:

Let x = length and y = width

Then, 2x+2y = 28 and xy = 48

Since 2x+2y = 28, y = 14 - x

So, x(14 - x) = 48

x2 - 14x + 48 = 0

(x-8)(x-6) = 0

x = 8 or x = 6

Hope this helps!

Answer 2

The dimensions are 6 ft by 8 ft

Explanation:

Area = l*w

Perimeter = 2(l+w)

48 = lw

28 = 2(l+w)

Divide by 2

28/2 = 2/2(l+w)

14 = l+w

Subtract w from each side

14-w =l+w-w

14-w =l

Substitute this into the first equation for area

48 = lw

48 = (14-w) *w

48 = 14w - w^2

Subtract 14w from each side

48 -14w = 14w-14w -w^2

-14w +48 = -w^2

Add w^2 to each side

w^2 -14w +48 = -w^2+w^2

w^2 -14w +48 =0

Factor

(w-6) (w-8) = 0

Using the zero product property

w-6=0 w-8=0

w=6 w=8

So the width could be 6 or the width could be 8

If the width is 6

14-w =l

14-6 =l

8 =l

The length is 8

If the width is 8

14-w =l

14-8 =l

8 =l

The length is 6