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3 years ago

the perimeter of a rectangle field is 340 yd. the length is 30 yd longer than the width. find the dimensions

Mathematics

2 answers:

zvonat [6]3 years ago

8 0

Lets assume:

Width=x

Length=x+30

2x+2(x+30)=340

2x+2x+60=340

4x+60=340

4x=280

x=70

Plugging x in:

Width=70

Length=100

Ivahew [28]3 years ago

4 0

Let l: length, w=width
2(l+w)=340, l=w+30
l+w=170
w+30+w=170
2w=140
w=70 yd.
l=70+30=100 yd
So the length of the rectangle field is 100 yd. and the width of the rectangle field is 70 yd. Hope it help!

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Putting the values we get

$\frac{4}{A} = \frac{5}{B} = \frac{10}{16}\\\frac{4}{A} = \frac{5}{B} = \frac{5}{8}\\Now\\\frac{4}{A} = \frac{5}{8}\\\frac{A}{4} = \frac{8}{5}\\A = \frac{8}{5} * 4\\A = \frac{32}{5}\\And\\\frac{5}{B} = \frac{5}{8}\\\frac{B}{5} = \frac{8}{5}\\B = 8$

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