The perimeter of a rectangle is twice the length plus twice the width so half the perimeter is the length plus the width.
If the perimeter is 12 m, then the length plus the width must equal 6. The dimensions could therefore be 1 m x 5 m, 2 m x 4 m, 3 m x 3 m (although this would be a square), 4 m x 2 m or 5 m x 1 m.
You forgot to put in the length of the rectangle's perimeter.
a rectangle has a perimeter of 72m. If the length is 20 m longer than the width find its dimensions?
7 x 19 cm
Length = 9 Width = 9 Your rectangle is a square.
If the given dimensions are the length and width of a rectangle then its perimeter is 15.6+4.5+15.6+4.5 = 40.2
A rectangle has two dimensions - length and width. Only if both dimensions are doubled, then the perimeter will be doubled.
You forgot to put in the length of the rectangle's perimeter.
a rectangle has a perimeter of 72m. If the length is 20 m longer than the width find its dimensions?
The dimensions of the rectangle will then work out as 14 cm by 10 cm because the perimeter is 14+10+14+10 = 48 cm
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?
7 x 19 cm
Length = 9 Width = 9 Your rectangle is a square.
A rectangle by definition has two pairs of sides with equal length. Since perimeter equals the length of all the sides. The equation for the perimeter of a rectangle could be thought of as: 2L + 2W = P Where L represents the length of one side of the rectangle and W represents the length of the adjacent (next to) side of the rectangle. If you know the length of one side and the perimeter, plug those values in as L and P and then solve for W. That will give you L and W which are the dimensions of the rectangle.
If the given dimensions are the length and width of a rectangle then its perimeter is 15.6+4.5+15.6+4.5 = 40.2
The dimensions of a rectangle are the length and the width. With these two measurements , the area of the rectangle can be calculated : Area = length x width. The perimeter can also be found : Perimeter = (2 x length) + (2 x width).
No, it is not possible for a rectangle to have a perimeter of 46 and an area of 42 simultaneously. For a rectangle, the perimeter ( P ) is given by ( P = 2(l + w) ), and the area ( A ) is ( A = l \times w ), where ( l ) is the length and ( w ) is the width. Solving these equations shows that the dimensions needed for these values are inconsistent, meaning no such rectangle exists.
If you increase the rectangle's length by a value, its perimeter increases by twice that value. If you increase the rectangle's width by a value, its perimeter increases by twice that value. (A rectangle is defined by its length and width, and opposite sides of a rectangle are the same length. The lines always meet at their endpoints at 90° angles.)