Assuming the area is rectangular, the area is width * length = 2*1 = 1 square unit.
32.6 (rounded to 1 dp.) Method; Area = length x width Rearranging ... Length = Area / width Length = 228/7 = 32.6
width 1, length 10 width 2, length 5
area = length * width so...300 square feet 300 = 1 * ? your answer is 300.
The area of a rectangle is the length times the width. 3 is the length and 1 is the width, 3 times 1 is 3 so therefore the area is 3.
It depends on the shape. To find the area of a square, multiply two sides (length x width), or find the square of a side (s2). To find the area of a rectangle, multiply length times width. The area of a triangle is 1/2 base x height, or (base x height)/2.
the width is 1 and the length is 2. 2 multipleyed by 1 equals 2...
The length is 44 1/2 feet, and the width is 39 1/2 feet
(Width 1 + Height 1 + Width 2 + Height 2)* Length
Perimeter = 2 lengths + 2 widths 1/2 perimeter = length + width Length = 1/2 perimeter - width Area = length x width That's (1/2 perimeter - width) x (width) .
32.6 (rounded to 1 dp.) Method; Area = length x width Rearranging ... Length = Area / width Length = 228/7 = 32.6
To find the area of a rectangle, multiply the length times the width.
There is no sure way to find the height of a rectangular prism with just the length and width. You need some other defined variable like the area. The height can range from 1 to infinity and never affect the base, length, and width.
width 1, length 10 width 2, length 5
area = length * width so...300 square feet 300 = 1 * ? your answer is 300.
The area of a rectangle is the length times the width. 3 is the length and 1 is the width, 3 times 1 is 3 so therefore the area is 3.
The surface area is length times width plus length. Then you find the square root of the width divided by two and then squared. You add this to the height squared plus the width. The width is multiplied by the square root of 1/2 squared plus the height squared.
Information about area is not enough to determine its dimensions. You can, for example, double its length and halve its width without altering the area.