Geometry. The perimeter is a rectangle is 62 in.?

Geometry. The perimeter is a rectangle is 62 in. If the length of rectangle is 1in more than twice its width, what are the dimensions of the rectangle?

L = 2w + 12(2w + 1) + 2w = 626w = 60w = 10l = 2w + 1 = 2*10 + 1l = 21.

1) perimeter = 2lenght + 2width2) your condition: length = 2width + 1And now: 62 = 2(2width + 1) + 2width 62 = 6width +2 60 = 6width ==> width = 10And finally length = 2(10) + 1 = 21

I googled example of perimeter of rectangle and this came up. Exactly answers your question.































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Answers with SolutionsQns. 1:Length = l + 1Width = lPerimeter of Rectangle = 2 Lengths + 2 Widths = 2 ( l + 1 ) + 2 ( l ) = 4l + 24l + 2 = 624l = 60l = 60 / 4 = 15 in.Substitute l = 15 in. into Length & Width,Therefore, Length = 15 + 1 = 16 in.Width = 15 in.Hope This Solutions Will Help In Answering Your Questions G0n9 G0n9â„¢Mathematician