PhantomPac
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Sagot :

mhanifa

Answer:

  • 14 m

Step-by-step explanation:

Let the shorter side be x

Then the longer side is x + 1

Perimeter is:

  • P = 2( x + x + 1) = 2(2x + 1) = 4x + 2

The value of perimeter is at most 60 meters. P ≤ 60.

Solve the inequality:

  • 4x + 2 ≤ 60
  • 4x ≤ 58
  • x ≤ 58/4
  • x ≤ 14.5

Since x is the integer, its greatest value is 14 m

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[tex]{ \large{ \red{➾ \: \: \underline {\bf{ \underline{Let \: \: the \: \: width \: \: of \: \: rectangle \: \: be \: \: y}}}}}}[/tex]

[tex]{ \large{ \blue{➾ \: \: \underline {\bf{ \underline{Then \: \: the \: \: length \: \: of \: \: rectangle \: \: be \: \: (y + 1)}}}}}}[/tex]

[tex]{ \large{ \green{✪ \: \: \boxed{ \boxed{ \rm{Perimeter \: \: of \: \: rectangle = 2 \times (length + breadth)}}}}}}[/tex]

[tex]{ \large{ \pink{⇛ \: \: \sf{Perimeter = 2 \times (y + y + 1)}}}}[/tex]

[tex]{ \large{ \orange{⇛ \: \: \sf{Perimeter = 4y + 2}}}}[/tex]

[tex]{ \large{ \purple{ ✒ \: \: \: \tt{Here \: \: given = \: Perimeter \: \: is \: \: at \: \: most \: \: 60 \: m}}}}[/tex]

[tex]{ \large{ \red{ ⇝ \: \bf{So \: \: perimeter \leqslant 60}}}}[/tex]

[tex]{ \large{ \green{ \underline{ \rm{Now \: \: we \: \: have \: \: to \: \: solve \: \: inequality}} \colon}}}[/tex]

[tex]{ \large{ \blue{➠ \: \: \bf{4y + 2 \leqslant 60}}}}[/tex]

[tex]{ \large{ \orange{➠ \: \: \bf{4y \leqslant 58}}}}[/tex]

[tex]{ \large{ \purple{➠ \: \: \bf{y \leqslant { \cancel\dfrac{58} {4 }^ { \: \: {14.5}} {}} }}}}[/tex]

[tex]{ { \blue{➠ \: \: \bf{y(Width \: \: of \: \: rectangle ) \: \: \leqslant 14.5 \: \: \: { \underline{ \underline{(But \: \: y \: \: is \: \: an \: \: integer)}}}}}}} [/tex]

[tex]{ \large{ \rm{ \green{ ∴ \: \: \: \: { \boxed{ \boxed{\bf{y \: (Width \: \: of \: \: rectangle ) \: \: greatest \: \: value \: = 14 \: \: m \: \: \:{ \underline{ \underline{}}}}}}} }}} \: \: { \large{ \green{ \rm{Ans.}}}}}[/tex]

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