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AǫᴜᴀWɪᴢ

[tex]\quad \huge \quad \quad \boxed{ \tt \:Answer }[/tex]

[tex]\qquad \tt \rightarrow \: length = 11 \: \: in[/tex]

[tex]\qquad \tt \rightarrow \: width = 4 \: \: in[/tex]

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[tex] \large \tt Solution \: : [/tex]

Let the measures be :

  • width = x

  • length = x + 7

[tex] \textsf{\large Area of rectangle :} [/tex]

[tex]\qquad \tt \rightarrow \: Area = length \sdot width [/tex]

[tex]\qquad \tt \rightarrow \: 44 = x \sdot(x + 7)[/tex]

[tex]\qquad \tt \rightarrow \: {x}^{2} + 7x = 44[/tex]

[tex]\qquad \tt \rightarrow \: {x}^{2} + 7x - 44 = 0[/tex]

[tex]\qquad \tt \rightarrow \: {x}^{2} + 11x - 4x - 44 = 0[/tex]

[tex]\qquad \tt \rightarrow \: x(x + 11) - 4(x + 11) = 0[/tex]

[tex]\qquad \tt \rightarrow \: (x + 11)(x - 4) = 0[/tex]

possible values of x = -11 and 4

[tex] \textsf{But the acceptable value of x = 4} [/tex]

[ length or width of rectangle can't be negative ]

[tex]\qquad \tt \rightarrow \: length = x + 7 = 4 + 7 = 11 \: \: in[/tex]

[tex]\qquad \tt \rightarrow \: width = x = 4 \: \: in[/tex]

Answered by : ❝ AǫᴜᴀWɪᴢ ❞

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