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The width of a rectangle is 3 inches less than its length, and the area is 88 square inches. What are the length and width of the rectangle?

User Tommy  Yu
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\large{ \tt{❃ \: EXPLANATION}} :

  • Remember - The formula to find out the area of a rectangle is l × w where l = length & w = width.

  • Let the length of a rectangle be ' l ' . We're provided - Width ( w ) = l - 3 & area of rectangle ( A ) = 88 square inches. We're asked to find out the length and width of the rectangle. Let's begin !


\large{ \tt{❁ \: SOLUTION}} :


\large{ \tt{♡ \: AREA \: OF \: RECTANGLE \: ( \: A \: ) = \bf l * w}}

  • Plug the values and then simplify :


\large{ \text{⟶ \: 88 = l(l - 3)}}


\large{ \bf{⟶ \: 88 = {l}^(2) - 3 l}}


\large{ \bf{⟶ {l}^(2) - 3l = 88 }}


\large{ \bf{⟶ \: {l}^(2) - 3l - 88 = 0}}


\large{ \bf{ ⟶{l}^(2) - (11 - 8)l - 88 = 0}}


\large{ \bf{⟶ {l}^(2) - 11l + 8l - 88 = 0}}


\large{ \bf{⟶ \: l(l - 11) + 8(l - 11) = 0}}


\large{ \bf{⟶ \: (l - 11)(l + 8) = 0}}

  • Either :


\large{ \bf{⟶ \: l - 11 = 0}}


\large{ \bf{⟶l = 11 \: in}}

  • Or :


\large{ \bf{⟶l + 8 = 0}}


\large{ \bf{⟶l = - 8 \: in}}

  • We found out the two values of length of given rectangle [ i.e 11 in & 8 in ]. Since the length can't be in negative , we consider length of the rectangle as ' 11 in ' .


\large{ \tt{❈ \: NOW, \: REPLACING \: VALUE}} :


\large{ \tt{⤑ \: Width(w) = l - 3 = 11 * 3 = \underline{ \tt{8 \: in}}}}


\underline{ \underline{ \large{ \tt{☄OUR \: FINAL \: ANSWER \: }}} }:


  • \large{ \boxed{\tt{Length = \underline{ 11 \: in}}}}


  • \large{ \boxed{ \tt{Width = \underline{8 \: in}}}}

☂ Yippie! we're done!

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User Robodisco
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