If length is "l" and width is "w", then the equation can be written as:
because we know that in a quadrilateral, the perimeter is the length plus the length plus the width plus the width, or 2 times the length plus 2 times the width.
To simplify 2l+2w=150, we can divide each side by two:
2l+2w=150 (150 divided by 2= 75)
Now we know that the length plus the width is 75 units.
We know that the length is twice the length is twice the width, or:
We can substitute this in for the length.
Now, we can divide each side by 3 to get the width:
3w=75 (75 divided by 3= 25)
The width=25 units
Now, we can figure out what the length is by putting in 25 for the width:
We can subtract each side by 25:
The length=50 units
We know this is correct because 50 (the length) is two times 25 (the width)
Length = 20 m and width = 9 m
If it's a rectangle, just minus the length from the perimeter twice and than divide what you have by 2. Width = (Perimeter - (length*2))/2
There seems to be a degree of difficulty about perimeter, which is just the measurement of the outline of the figure. It appears that your figure is a rectangle so the perimeter is merely twice (the sum of length and width, ie 150 units). If length exceeds width by 20 then length is 85 and width is 65.
228 feet is the perimeter.
the width equals 14 cm. and the length equals 31 cm.
length + width = ½ perimeter ie 24', so length = 24 - width length = 2 x width + 3, so 24 - width = 2 x width + 3 ie 21 = 3 x width so width = 7 and length = 17
The length is 14 cm
The dimensions of the rectangle are 21 units by 9 units
Designate the width by w. Then, from the problem statement, the length = 6 + 2w. The perimeter is twice the width plus twice the length, or: 2w + 2(2w + 6) = 36; or 2w + 4w + 12 = 36; or 6w = 24 or w = 4 meters; the length is 14 meters.
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
First of all, the area cannot be 40 cm since an area is 2-dimensional whereas the measure given is 1-dimensional. Next, the shape is not defined and so the width, length and perimeter are indeterminate. Even if you assume that the shape is a rectangle, the width, length and perimeter are indeterminate. On the basis that the length is greater than the width, all that you can say is that the length will be greater than or equal to sqrt(40) = 6.324555 (to 6 dp), the width will be 40 divided by that length and the perimeter will be twice their sum. Some examples: Length = 8, width = 5, perimeter = 26 Length = 10, width = 4, perimeter = 28 Length = 40, width = 1, perimeter = 82 Length = 4000, width = 0.01, perimeter = 8000.02 Length = 4,000,000, width = 0.00001, perimeter = 8,000,000.00002 and so on, without limit.
Width = 16 feet Length = 38 feet
length is 3 and width is 9
Length of rectangle is 5 inches and its width is 4 inches Check: 2*(5+4) = 18 inches which is its perimeter
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 perimeter of a standard- sized rectangular rug is 94 ft. the length is 3 ft more than the width. Find the length and the width.
Badly expressed! Do you mean "what is the area of a rectangle with a length that is 3 more than twice its width?" ? If you have a width, then presumably you have a length, in which case you presumably have a rectangle. But what do we want to know about this rectangle? You haven't said. Its area? Its perimeter? Its color? Its shoe size?
length=18cm width=10cm (you use a system of equations: 2L+2W=56, L=2W-2)
The length of the length will be 5 and the width will be 25.