Given: P = 150 and L = (5W + 3) Solve the 'L' equation for 'W': (L-3) = 5W ===> W = (L-3)/5 Perimeter = 2L + 2W = 2L + 2[ (L-3)/5 ] = 2L + (2L-6)/5 5 x perimeter = 10L + (2L-6) = 12L - 6 Perimeter = 150 = (12L-6)/5 ===> 750 = (12L - 6) ===> 756 = 12L ===> L = 756/12 = 63=============================== Check: L = 63, W = 12, Perimeter = 2(L+W) = 2(63+12) = 2(75) = 150. Yay !
21cm3
Perimeter of a rectangle is 2(l+b).Area of rectangle is 216 centimeter square.
A rectangle with a perimeter of 48 and 1 side equal to 5 would have its other sides lengths of 5, 19, 19. Such a rectangle would have an area of 95 sq/m. Your question is somewhat strange, as the "length" of a rectangle generally refers to the longer side - in this case the length would be 19 and the width 5.
L + W = P/2 = 13. 13 - 7 = 6, 6/2 = 3 which is the width, making the length 10.
if the perimeter is 46m, then side + length = 23 so we need two values, when added together are 23, with a difference of 11 6 + 17 = 23, (and have a difference of 11) therefore the length = 17m, width = 6m
a rectangle as a longer side because there's 48 inches for a rectangle but there's only 45 inches for a triangle. a rectangle as a longer side because there's 48 inches for a rectangle but there's only 45 inches for a triangle.
Length = 3.5 inches and Width = 2.5 inches Check: 2(3.5+2.5) = 12 inches
It is: 12 units in length
a rectangle has a perimeter of 72m. If the length is 20 m longer than the width find its dimensions?
shorter side = 10 longer side = 19
i don't know what length of rectangle is 7 cm longer than the width the perimeter of the rectangle is 46 cm .what is the length of the of the rectangle ?
The perimeter actually doesn't tell you anything about the breakdown betweenthe length and width. All we can tell about this one is that the sum of (length+width)has to be 15 inches. That still leaves an infinite number of possible rectangles.If the perimeter is 30 inches and the length has to be the longer dimension, thenit can be anything more than 7.5 inches, with no upper limit.For example:Width = 1 millionth of an inchLength = 1 millionth of an inch less than 15 inchesPerimeter = 30 inches.
21cm3
Perimeter of a rectangle is 2(l+b).Area of rectangle is 216 centimeter square.
The perimeter actually doesn't tell you anything about the breakdown betweenthe length and width. All we can tell about this one is that the sum of (length+width)has to be 98 inches. That still leaves an infinite number of possible rectangles.If the perimeter is 196 inches and the length has to be the longer dimension, thenit can be anything more than 49 inches, with no upper limit.For example:Width = 1 millionth of an inchLength = 1 millionth of an inch less than 98 inchesPerimeter = 196 inches.
the width equals 14 cm. and the length equals 31 cm.
Length + width = half of perimeter = 37 in. 37 - 9 = 28, half of 28 = 14 which is the width, and the length is 23