Question

The length of a rectangle is four times its width. If the perimeter of the rectangle is 50 yd, find its area

Answer 1

100yd²

Explanation:

length=4x

width=x

perimeter=2(l+w)

50=2(4x+x)

50=2(5x)=10x

50=10x

x=5yd

width=5yd

length=20yd

area=length×width

=20×5

=100yd²

Answer 2

`\boxed{\red{100 \: \: {yd} ^(2)}}`

Explanation:

width = x

length = 4x

so,

perimeter of a rectangle

`p= 2(l + w) \\ 50yd = 2(4x + x) \\ 50yd= 2(5x) \\ 50yd= 10x \\ (50yd)/(10) = (10x)/(10) \\ x = 5 \: \: yd`

So, in this rectangle,

width = 5 yd

length = 4x

= 4*5

= 20yd

Now, let's find the area of this rectangle

`area = l * w \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 20 * 5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 100 {yd}^(2)`