w x 2w = 63. 2w^2 = 63 w^2 = 31.5 w = sqrt 31.5 = 5.6125 so L = 11.225 which gives area of 63.0003, which is near enough for Wiki!
From the statement of the problem, if w is the width, the area is 2w2 , the product of the width and the length, which is stated to be twice the width. Since 2w2 must be less than 50, w2 < 25, and the width must be less than 5 meters.
4 cm
9" x 4"
Let length be denoted by l, and width by w. Length is 4 ft more than twice the width => l = 2w + 4 five times the width is the same as twice the length increased by 10 feet. =>5w = 2l + 10 Solving the two equations we get l = 40 w = 18 So, the rectangle is 40 ft long and 18 ft wide.
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
From the statement of the problem, if w is the width, the area is 2w2 , the product of the width and the length, which is stated to be twice the width. Since 2w2 must be less than 50, w2 < 25, and the width must be less than 5 meters.
3 and 7
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?
Factors of 28 include 7 x 4. 7 is 1 less than twice 4...
Length = 20 m and width = 9 m
lenght=15, width= 10
12 meters
15
154 Divided by8
Area of a rectangle is length times width, therefore if the length is two feet longer than the width then the length is 9 and the width is 7, when they are multiplied you get 63
4 cm
Area of the rectangle = 36 =Length * Width Assume Width= a Hence Length = 4*a =4a Area = a*4a =36 Therefore, a*a =9 Hence a =3 =Width