Suppose the width is W yards.
Then twice the width is 2W yards
3 yards less that twice the width is (2W - 3) yards.
That is, the length is (2W - 3) yards
So, the area = Length *Width = (2W - 3)*W which is known to be 27
Therefore 2W2 - 3W = 27
or 2W2 - 3W - 27 = 0
so (W + 3)*(2W - 9) = 0
so W = -3 or W = 4.5
Since W cannot be negative, W = 4.5 yards and then L = 6 yards.
Suppose the width is W yards.
Then twice the width is 2W yards
3 yards less that twice the width is (2W - 3) yards.
That is, the length is (2W - 3) yards
So, the area = Length *Width = (2W - 3)*W which is known to be 27
Therefore 2W2 - 3W = 27
or 2W2 - 3W - 27 = 0
so (W + 3)*(2W - 9) = 0
so W = -3 or W = 4.5
Since W cannot be negative, W = 4.5 yards and then L = 6 yards.
Suppose the width is W yards.
Then twice the width is 2W yards
3 yards less that twice the width is (2W - 3) yards.
That is, the length is (2W - 3) yards
So, the area = Length *Width = (2W - 3)*W which is known to be 27
Therefore 2W2 - 3W = 27
or 2W2 - 3W - 27 = 0
so (W + 3)*(2W - 9) = 0
so W = -3 or W = 4.5
Since W cannot be negative, W = 4.5 yards and then L = 6 yards.
Suppose the width is W yards.
Then twice the width is 2W yards
3 yards less that twice the width is (2W - 3) yards.
That is, the length is (2W - 3) yards
So, the area = Length *Width = (2W - 3)*W which is known to be 27
Therefore 2W2 - 3W = 27
or 2W2 - 3W - 27 = 0
so (W + 3)*(2W - 9) = 0
so W = -3 or W = 4.5
Since W cannot be negative, W = 4.5 yards and then L = 6 yards.
Suppose the width is W yards.
Then twice the width is 2W yards
3 yards less that twice the width is (2W - 3) yards.
That is, the length is (2W - 3) yards
So, the area = Length *Width = (2W - 3)*W which is known to be 27
Therefore 2W2 - 3W = 27
or 2W2 - 3W - 27 = 0
so (W + 3)*(2W - 9) = 0
so W = -3 or W = 4.5
Since W cannot be negative, W = 4.5 yards and then L = 6 yards.
If the length of a rectangle is twice its width and it has a perimeter of 48, then the rectangle is 16 in length and 8 in width.
The dimensions are 16 units by 8 units of mearsurement
9" x 4"
Given: p = 26cm and 2x width - 5cm = length p = 2(a+b) = 26cm; a+b = 13cm 2x6-5=7 The sides of the rectangle are: width 6cm and length 7cm
this is a funky question... 8x10 kilometers
If the length of a rectangle is twice its width and it has a perimeter of 48, then the rectangle is 16 in length and 8 in width.
The dimensions of the rectangle are 21 units by 9 units
width=15 length=35
A rectangle has two dimensions, length and width. You haven't said what the width is. The perimeter is the distance around the rectangle. Imagine going round it. You would go 42 inches, then a width, then another 42 inches, then another width. In general, the perimeter of a rectangle is twice its length plus twice its width.
The dimensions are 16 units by 8 units of mearsurement
The length of a rectangle is twice its width. If the perimeter of the rectangle is , find its area.
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.)
It has 2 dimensions which are length and width
9" x 4"
4cm by 13cm
They are also known as the "dimensions" of a rectangle.
2x4