Length is usually the longest dimension.
L + W = P/2 = 13. 13 - 7 = 6, 6/2 = 3 which is the width, making the length 10.
To find the area of a rectangle, you multiply its length by its width. In this case, 10 meters by 7 meters equals an area of 70 square meters.
10 inches
The perimeter would be 28 meters.
There are an infinite number of possibilities. Let A metres be any positive length and let B be 10/A metres. Then the rectangle of size A*B will have an area = A*B = 10 sq metres. Furthermore, there is no requirement for the shape to be a rectangle. It could be a circle, ellipse, triangle etc or a completely irregular shape whose length and width are difficult to define.
Diagonal = 10 meters.
we know that area of rect= l* b area = 11* 10 =110 cm^2
The perimeter of a rectangle is given by:P = 2l + 2w, where P is the perimeter of the rectangle, l is the length of the rectangle, and w is the width of the rectangle.P = 2(15) + 2(10) = 50 meters.
A(rect) = w * l = 12ft*10ft = 120ft2Length * width = area10 * 12 = 120
area you have to multiply the width by the length so if you do 10*7 you will get 70. the answer is 70 meters squared.
The diagonal is 15.620 meters.
Assume the area of the rectangle is 70 squarecentimeters.Convert the length to centimeters : 1 meter = 100 cm : 10 meters = 1000 cmArea of a rectangle = length x width.70 = 1000 x width : width = 70/1000 = 0.07 cmPerimeter of a rectangle = 2 (length + width) = 2(1000 + 0.07) = 2 x 1000.07 = 2000.14 cm = 2000.14/100 = 20.0014 meters.
You multiple one side by the other, so 10 x 10 = 100 square meters.
To find the area of a rectangle, you multiply the length by the width. In this case, the length is 10 meters and the width is 9 meters. So, the area would be 10 meters x 9 meters = 90 square meters.
42.2 P = 2L + 2W P = 7.6 + 34.6
To find the length of the diagonal in a rectangle, use the formula for calculating the hypotenuse of a right triangle: a² = b² + c². a² = 3² + 10² a² = 9 + 100 a² = 109 a = 10.44 m
The dimensions of a rectangle with an area of 400 square meters can vary, as multiple pairs of length and width can produce this area. For example, if the length is 20 meters, the width would be 20 meters (making it a square). Alternatively, if the length is 40 meters, the width would be 10 meters. Thus, any combination of length and width that multiplies to 400 will satisfy the area requirement.