Assuming that the area given is actually in squareinches, this can be solved as follows: Call the width inches w. Then, from the problem statement, the length is w + 3 inches. The area of any rectangle is the product of its length and width, so that:
w(w+3) = 154; or w2 + 3w - 154 = 0, or (w + 14)(w - 11) = 0. This positive value of w for which this equation is true is 11 inches.
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
Area=Length*Width A=14 Inches*2 Inches A= 28 square inches or 28 in2
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)
Length of rectangle is 5 inches and its width is 4 inches Check: 2*(5+4) = 18 inches which is its perimeter
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.
To find the area of a rectangle, multiply the length times the width. The result will be in square inches.
Area=Length*Width A=14 Inches*2 Inches A= 28 square inches or 28 in2
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.