The perimeter of a rectangle = 2L + 2W So, from your problem we know that L = W + 27 Use this equation for the length and substitute in the perimeter equation, we get: 2(W+27) + 2W = 96 Now we rearrange and solve for W 2W + 54 + 2W = 96cm 4W + 54 -54 = 96 -54 4W = 42 W = 10.5cm, and using this value L = 10.5 + 27 = 37.5cm
Let the width of rectangle is w cm (Length is 12 cm more) so length is w + 12 Perimeter = 16 times of width 2 (length + width) = 16 * width Rest of the homework, do yourself.
length + width = 20/2 = 10 length = 7cm, width = 3cm
Length = 20 m and width = 9 m
The length could be 3 cm (width = 1 cm), with a perimeter of 8 cm, which is not more than 72 cm. Or it could be 6 cm (w = 2 cm, perimeter = 16 cm).
Do-it-in-your-head method: Length + width is half the perimeter which is 27 inches; 27 minus 6 is 21 and one-third of 21 is 7 which is the width. (Length = 2 x 7 + 6 = 20)
the length of a rectangle is 5 more then the width. Find the perimeter and the area of the rectangle
10x2
width=5 length=12
Let the width of rectangle is w cm (Length is 12 cm more) so length is w + 12 Perimeter = 16 times of width 2 (length + width) = 16 * width Rest of the homework, do yourself.
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?
You forgot to put in the length of the rectangle's perimeter.
w=8in l=13in
Perimeter = 2 lengths and 2 widths In your case length + width = 25cm If length is 5cm more than width then length = 15cm and width = 10cm
length + width = 20/2 = 10 length = 7cm, width = 3cm
Length = 20 m and width = 9 m
The dimensions of the rectangle are 21 units by 9 units
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