ellamai50
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A farmer wishes to enclose a rectangular plot with a wire fence. The width of the plot is 3 metres less than its length (x). The area enclosed by the fence is 378 square metres. Form an equation in x and hence find the length of the plot.


Help​

Sagot :

  • Length is x
  • Width be x-3

[tex]\\ \rm\rightarrowtail Area=Length\times Width[/tex]

[tex]\\ \rm\rightarrowtail x(x-3)=378[/tex]

[tex]\\ \rm\rightarrowtail x^2-3x=378[/tex]

[tex]\\ \rm\rightarrowtail x^2-3x-378=0[/tex]

[tex]\\ \rm\rightarrowtail x^2-21x+18x-378=0[/tex]

[tex]\\ \rm\rightarrowtail (x+18)(x-21)=0[/tex]

[tex]\\ \rm\rightarrowtail x=-18,21[/tex]

Take it positive

  • Length=21m

Let the length be x

Let the width be x-3

Now,

[tex] \red \dashrightarrow \mathcal{Area = Length×Breadth}[/tex]

  • Area is given i.e. 378 m²

[tex] \red \dashrightarrow \sf \: 378 = x \times (x - 3)[/tex]

[tex] \red \dashrightarrow \sf \: 378 = {x}^{2} - 3x[/tex]

[tex] \red \dashrightarrow \sf \: 378 - {x}^{2} + 3x = 0[/tex]

[tex] \red \dashrightarrow \sf \: - 378 + {x}^{2} - 3x = 0[/tex]

[tex] \red \dashrightarrow \sf \: {x}^{2} - 3x - 378= 0[/tex]

[tex] \red \dashrightarrow \sf \: {x}^{2} + 18x - 21x - 378= 0[/tex]

[tex] \red \dashrightarrow \sf \: x \times (x + 18) - 21(x + 18)= 0[/tex]

[tex] \red \dashrightarrow \sf \: (x + 18) (x - 21)= 0[/tex]

[tex] \red \dashrightarrow \sf \: x + 18= 0 \\ \red \dashrightarrow \sf \: x - 21= 0[/tex]

[tex] \red \dashrightarrow \sf \: x = - 18 \\ \red \dashrightarrow \sf \: x = 21[/tex]

  • Since the dimension cannot be in negative, We will consider x is 21

[tex] \colorbox{lightyellow}{length = x = 21}[/tex]

[tex] \colorbox{lightyellow}{breadth= x - 3= 21 - 3 = 18}[/tex]

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