Not enough information. You need to know how many sides and if there are any other relationships between the sides (like the length is 2.5 times the width, for example). One possible solution is width = 5 and length = 12.5 (assuming a rectangle).
Perimeter = 15+15+35+35 = 100 cm
If the shape is a rectangle, then Perimeter = 2(Length + Width) So Width = Perimeter/2 - Length
perimeter = 2 x (length + width) ⇒ length + width = perimeter ÷ 2 ⇒ length = (perimeter ÷ 2) - width So to find the length of the given rectangle, subtract the width from half the perimeter is how to do it. IF you meant "What is the length..." then: length = perimeter ÷ 2 - width = 40 ÷ 2 - 2 in = 20 - 2 in = 18 in
Length + length + width + width = perimeter. Length + width x 2 = perimeter
Perimeter = 2 * width + 2 * length, so rearranging --> width = (Perimeter / 2) - length
Perimeter = 15+15+35+35 = 100 cm
35 mm
If the shape is a rectangle, then Perimeter = 2(Length + Width) So Width = Perimeter/2 - Length
35 and 28
perimeter = (2*length) + (2*width) length = 2*width so perimeter = (2*2*width) + 2*width = 6*width perimeter = 48 so you can figure out the width and length
perimeter = 2 x (length + width) ⇒ length + width = perimeter ÷ 2 ⇒ length = (perimeter ÷ 2) - width So to find the length of the given rectangle, subtract the width from half the perimeter is how to do it. IF you meant "What is the length..." then: length = perimeter ÷ 2 - width = 40 ÷ 2 - 2 in = 20 - 2 in = 18 in
Length + length + width + width = perimeter. Length + width x 2 = perimeter
Perimeter = 2 lengths + 2 widths 1/2 perimeter = length + width Length = 1/2 perimeter - width Area = length x width That's (1/2 perimeter - width) x (width) .
width = (perimeter-2*length)/2
Perimeter = 2 * width + 2 * length, so rearranging --> width = (Perimeter / 2) - length
Perimeter = 24 inches Area = 35 square inches
To draw a rectangle with an area of 35 square units and a perimeter of 35 units, you can use the formulas for area (A = length × width) and perimeter (P = 2(length + width)). Let the length be ( l ) and width be ( w ). From the area, you get ( lw = 35 ), and from the perimeter, ( 2(l + w) = 35 ) or ( l + w = 17.5 ). Solving these equations simultaneously, you can find suitable dimensions, such as ( l = 10 ) units and ( w = 3.5 ) units or ( l = 7 ) units and ( w = 5 ) units.