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malfutka [58]
3 years ago
12

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:

<em>width</em><em> </em><em>=</em><em> </em><em>x</em>

<em>length</em><em> </em><em>=</em><em> </em><em>4</em><em>x</em>

<em>so</em><em>,</em><em> </em>

<em>perimeter</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>rectang</em><em>l</em><em>e</em><em> </em>

<em>p= 2(l + w) \\ 50yd = 2(4x + x) \\  50yd= 2(5x) \\  50yd= 10x \\  \frac{50yd}{10}  =  \frac{10x}{10}  \\ x = 5 \:  \: yd</em>

<em>So</em><em>,</em><em> </em><em>in</em><em> </em><em>this</em><em> </em><em>rectangle</em><em>,</em>

<em>widt</em><em>h</em><em> </em><em>=</em><em> </em><em>5</em><em> </em><em>yd</em>

<em>length</em><em> </em><em>=</em><em> </em><em>4</em><em>x</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>4</em><em>*</em><em>5</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>=</em><em> </em><em>2</em><em>0</em><em>y</em><em>d</em>

<em>Now</em><em>,</em><em> </em><em>let's</em><em> </em><em>find</em><em> </em><em>the</em><em> </em><em>area</em><em> </em><em>of</em><em> </em><em>this</em><em> </em><em>rectangle</em>

<em>area = l \times w \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   = 20 \times 5 \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: = 100 {yd}^{2}</em>

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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