There are different sizes: commonly around 7 mm.
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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.
It is 22 in by 27 in. ------------------------------------------- How it is solved: The rectangle is 5 in longer than its width, so the dimensions can be given in terms of its width: Its dimensions are width by (width+5) In removing 1 inch squares from the corners, he flaps are going to be (width-2) and (width+5-2) = (width-3) long. Thus the box formed will have volume: volume = (width - 2) in × (width + 3) in × 1 in → volume = width² + width - 6 in³ But we are told this is 500 in³; thus: width² + width - 6 in³ = 500 in³ → width ² + width - 506 = 0 → (width + 23)(width - 22) = 0 → width = -23 or 22 As a length cannot be negative, the width must be 22 in which means the original rectangle is 22 in by 22+5 in = 27 in
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Area = Length * Width 24 = 6 * Width so width = 4m