Question

The perimeter of a rectangle is 84 cm. Twice the width is 9 less than the length. Find the dimensions of the rectangle

Answer 1

`\mathrm{Let\ }l\ \mathrm{and}\ w\ \mathrm{be\ the\ length\ and\ width\ of\ the\ rectangle\ respectively.}\\\mathrm{According\ to\ the\ question,}\\2w-9=l\\\mathrm{or, }\ 2w=l+9.........(1)\\\mathrm{We\ know,}\\\mathrm{Perimeter\ of\ rectangle}=2(l+w)=2l+2w\\\mathrm{From\ equation(1),}\\\mathrm{Perimeter\ of\ rectangle}=2l+(l+9)=3l+9\\\mathrm{or,\ }84=3l+9\\\mathrm{or,\ }3l=75\\\mathrm{or,}\ l=25cm\\\mathrm{Using\ equation(1),\ }2w=l+9=25+9\\\mathrm{or,\ }2w=34\\\mathrm{or,\ }w=17cm`

`\mathrm{So\ the\ length\ of\ rectangle\ is\ 25cm\ and\ the\ width\ is\ 17cm.}`

Answer 2

Width = 17 cm Length = 25 cm

Explanation:

Perimeter is found by taking 2W + 2L

It is stated that twice the width is 9 less than the length. The means that

L = 2W - 9

So substitute this for the L

P = 2W + 2L

84 = 2W + 2(2W-9)

84 = 2W + 4W - 18

84 = 6W - 18

102 = 6W

17 = W

L = 2(17) - 9 = 34-9 = 25