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3 years ago

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

Mathematics

2 answers:

Elden [556K]3 years ago

8 0

Answer:

$\boxed{\red{100 \: \: {yd} ^{2}}}$

Step-by-step explanation:

width = x

length = 4x

so,

perimeter of a rectangle

$p= 2(l + w) \\ 50yd = 2(4x + x) \\ 50yd= 2(5x) \\ 50yd= 10x \\ \frac{50yd}{10} = \frac{10x}{10} \\ x = 5 \: \: yd$

So, in this rectangle,

width = 5 yd

length = 4x

= 4*5

= 20yd

Now, let's find the area of this rectangle

$area = l \times w \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 20 \times 5 \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 100 {yd}^{2}$

SpyIntel [72]3 years ago

6 0

Answer:

100yd²

Step-by-step 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

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