To find the perimeter you have to 2 x L + 2 x W, 13 being you L and 15 your W. 13 x 2=26, 15 x 2=30. 30+26 is 56!
Your answer would be 26.......
65 yards
And the question is ...
perimeter = 2l + 2w l = 9 yards w = 12 yards perimeter = 2*9 yards + 2*12 yards = 18 + 24 = 42 yards
37
16 feet
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The perimeter of a rectangle of dimensions 50 yards x 100 yards is 300 yards.
P=2l+2w, so the perimeter is 21 yards.
If those are the dimensions of a rectangle then its perimeter is 2*(5+3)=16 yards
the perimeter of a rectangle is 700 yards. what are the dimensions of the rectangle if the lenght is 80 yards more than the width?
65 yards
And the question is ...
The dimensions are length 50 yards and width 25 yards
Field = 5y x1y where y depicts the measurement and 5:1 is the ratio. The perimeter of the field = 325 yards, assuming that opposite sides of the field are of equal length. Therefore assume that the perimeter of the field = 1y+1y+5y+5y =12y Set 12y = 325 yards to get y=27.0833 yards Therefor the field is 135.4167 yards x 27.0833 yards.
Length + width = half of perimeter so width = p/2 - length ie (22-12) 10 yards.
perimeter = 2l + 2w l = 9 yards w = 12 yards perimeter = 2*9 yards + 2*12 yards = 18 + 24 = 42 yards
37