Area=Length*Width
A=14 Inches*2 Inches
A= 28 square inches or 28 in2
Width = 9 inches Length = 30 inches
The dimensions are: length = 15 inches and width = 7 inches Check: 15*7 = 105 square inches
The perimeter of a rectangle = 2 x (length + width) = 2 x (12 + 8) = 2 x 20 = 40 inches
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)
Using Pythagoras' theorem it is about 10.81665383 inches.
Length of rectangle is 5 inches and its width is 4 inches Check: 2*(5+4) = 18 inches which is its perimeter
Let the width of the rectangle be represented by "w" inches. Since the length is twice the width, it can be expressed as "2w" inches. The formula for the perimeter of a rectangle is P = 2(l + w), where P is the perimeter, l is the length, and w is the width. Substituting the given values into the formula, we get 48 = 2(2w + w). Simplifying, we find that 48 = 6w. Solving for w, we find that the width of the rectangle is 8 inches, and the length is 16 inches.
7
Width = 9 inches Length = 30 inches
The dimensions are: length = 15 inches and width = 7 inches Check: 15*7 = 105 square inches
The perimeter of a rectangle = 2 x (length + width) = 2 x (12 + 8) = 2 x 20 = 40 inches
10 in. A+LS
To find the area of a rectangle, multiply the length times the width. The result will be in square inches.
To find the area of a rectangle, multiply the length times the width.
Length 7 inches and width 4 inches because 7*4 = 28 square inches
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)
To find the diagonal length of a rectangle use Pythagoras' theorem for a right angle triangle.