about 3/4 of an inch or 2 cm
Perimeter equals two times quantity length plus width, or 2(l+w), so: 2(4x+4x-2)=12 2(8x-2)=12 8x-2=6 8x=8 x=1 Plug 1 in for x Length: 4(1)=4 units Width: 4(1)-2=2 units
Area = length * width width = 4*102 length = 2*103 Area = 4*102*2*103 = 8*105
about 3/4 of an inch or 2 cm
2. Think about it. There are 4 sides. Two for width and two for length. And 2+2=4. Right?!
perimeter = 2 x (width + length) area = width x length If area = perimeter then: width x length = 2 x width + 2 x length width x length - 2 x length = 2 x width length x (width - 2) = 2 x width length = 2 x width / (width - 2) So any rectangle with length and width that satifies the above will work, for example: width = 2.5, length = 10, area: 2.5 x 10 = 25; perimeter: 2x(2.5+10) = 2x12.5 = 25 width = 3, length = 6, area: 3x6 = 18; perimeter: 2x(3+6) = 2x9 = 18 width = 4, length = 4, area: 4x4 = 16; perimeter: 2x(4+4) = 2x8 = 16 By making the length dependent on the width, the width must be at least 2. Once the 4x4 square has been reached, the width becomes larger than the length, but the pair will also be given if the width is the smaller dimension, for example, a width of 6 gives a length of 2 x 6 / (6 - 2) = 12 / 4 = 3, which matches the length of 6 being given by a width of 3.
Perimeter of a rectangle is (2 x Length) + (2 x Width) (2 x length)+(2 x width) = 18 length = 1+width 2(1+w) + (2w) = 18 2+2w+2w=18 2+4w=18 4w=16 w=4 l=1+4=5 Width = 4 Length = 5
Perimeter = length + width + length + width = 2 x (length + width) Given: perimeter = 22in length = width + 3in Thus 22 = 2 x (width + 3 + width) 11 = 2 x width + 3 8 = 2 x width 4 = width So the width is 4in.
About 4 1/2 inches long, and is about 2 1/2 inches width wise.
The width of an adult thumb could be somewhere between 2-4 cm.
Important to note are these formulae: Perimeter_of_rectangle = 2 x (length + width) Area_of_rectangle = length x width So if the perimeter and area are known, then: 2 x (length + width) = perimeter => length + width = perimeter / 2 => length = perimeter / 2 - width length x width = area => (perimeter / 2 - width) x width = area (substituting for length given above) => perimeter / 2 x width - width2 = area => width2 - perimeter / 2 x width + area = 0 which is a quadratic and can be solved either by factorization or by using the formula: width = (perimeter / 2 +/- sqrt(perimeter2 / 4 - 4 x area)) / 2 = (perimeter +/- sqrt(perimeter2 - 16 x area)) / 4 This will provide two values for the width. However, each of these values is the length for the other, so the larger value is the length and the smaller value is the width. Sometimes only 1 value will be found for the width above. In this case, the rectangle is actually a square which means that the length and width are both the same. Examples: 1. perimeter = 6, area = 2 width2 - perimeter / 2 x width + area = 0 => width2 - 6 / 2 x width + 2 = 0 => width2 - 3 x width + 2 = 0 => (width - 2) x (width - 1) = 0 => width = 2 or 1. So the length is 2 and the width is 1. 2. perimeter = 12, area = 9 width2 - perimeter / 2 x width + area = 0 => width2 - 12 / 2 x width + 9 = 0 => width2 - 6 x width + 9 = 0 => (width - 3)2 = 0 => width = 3 So the rectangle is a square with both length and width of 3.
about 3/4 of an inch or 2 cm
Perimeter equals two times quantity length plus width, or 2(l+w), so: 2(4x+4x-2)=12 2(8x-2)=12 8x-2=6 8x=8 x=1 Plug 1 in for x Length: 4(1)=4 units Width: 4(1)-2=2 units
Area=Length * Width. if length =4 and width =5...area is 20. if doubled....(4*2)*(5*2)=A*4. When length and width double, area is multiplied by 4.
Area = length * width width = 4*102 length = 2*103 Area = 4*102*2*103 = 8*105
2. Think about it. There are 4 sides. Two for width and two for length. And 2+2=4. Right?!
Length = 4, Width = 2