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.......
Let the width be ( w ) yards. Then, the length is ( 3w + 4 ) yards. The perimeter of a rectangle is given by ( 2(\text{length} + \text{width}) ), so we have the equation ( 2((3w + 4) + w) = 472 ). Simplifying gives ( 2(4w + 4) = 472 ), leading to ( 4w + 4 = 236 ) and ( 4w = 232 ), resulting in ( w = 58 ) yards. Thus, the dimensions are a width of 58 yards and a length of ( 3(58) + 4 = 178 ) yards.
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
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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?
Let the width be ( w ) yards. Then, the length is ( 3w + 4 ) yards. The perimeter of a rectangle is given by ( 2(\text{length} + \text{width}) ), so we have the equation ( 2((3w + 4) + w) = 472 ). Simplifying gives ( 2(4w + 4) = 472 ), leading to ( 4w + 4 = 236 ) and ( 4w = 232 ), resulting in ( w = 58 ) yards. Thus, the dimensions are a width of 58 yards and a length of ( 3(58) + 4 = 178 ) yards.
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