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The width of a rectangle is 12 cm less than the length. The perimeter is 156 cm. Find the width and length.

Sagot :

[tex]w-the\ width\\l-the\ length\\\\w=l-12\\\\the\ perimeter\ of\ rectangle:P=2(w+l)\ and\ P=156cm\\\\2(w+l)=156\to\ put\ w=l-12\ to\ equation:\\\\2(l-12+l)=156\\2(2l-12)=156\ \ \ \ |divide\ both\ sides\ by\ 2\\2l-12=78\ \ \ \ \ |add\ 12\ to\ both\ sides\\2l=90\ \ \ \ \ |divide\ both\ sides\ by\ 2\\l=45\\\\w=45-12=33\\\\Solution:the\ width\ equal\ 33cm\ and\ the\ length\ equal\ 45cm.[/tex]
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