Answer:

Solution:-

Step-by-step explanation:

Given:-

• Length and breadth of the plot are in the ratio = 11:4
• At the rate of Rs.100 per meter it will cost the village panchayat Rs.75,000 to fence the plot.

To find:-

• Dimensions of the plot

Required Formula:-

• Perimeter of rectangle = 2(l + b)

Solution:-

Where,

• length = 11x meter
• breadth = 4x meter

Then,

[tex]\leadsto \large \green{Perimeter=2(11x+4x)}[/tex]

[tex]\longmapsto \large \blue{Perimeter=2(15x)}[/tex]

[tex]\leadsto \large \dag \underline{ \boxed{Perimeter=30x}}[/tex]

Now,

• Finding the value of x by putting the value of cost of fencing and perimeter mutilplied by rate of per meter

[tex]\longmapsto \red{75000 = 30x \times 100}[/tex]

[tex]\longmapsto \blue{ \frac{750 \cancel{00}}{1 \cancel{00}} = 30x}[/tex]

[tex]\longmapsto \purple{750 = 30x}[/tex]

[tex]\longmapsto \large \green{ \frac{ \cancel{750} {}^{25} }{ \cancel{30}} = x }[/tex]

[tex]\longmapsto \dag{ \underline{\boxed{x = 25}}} \blue{\bigstar}[/tex]

Therefore,

• Length = 11 × 25 = 275 m

• Breadth = 4 × 25 = 100 m

Hence,

The length of the plot is 275 m and breadth of plot is 100 m.

[tex] \underline{\rule{200pt}{9pt}}[/tex]

Verified answer

Given :

• Cost of fencing = Rs.75000
• Rate of fencing = Rs.100 per m
• Ratio of Dimensions = 11:4

[tex] \\ \\ [/tex]

To Find :

• Find the Dimensions

[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]

SolutioN :

[tex] \maltese \; {\underline{\underline{\sf{ \; Formula \; Used \; :- }}}} [/tex]

• [tex] {\underline{\boxed{\pmb{\sf{ Perimeter {\small_{(Rectangle)}} = 2 ( Length + Breadth ) }}}}} [/tex]

[tex] \\ \\ [/tex]

[tex] \maltese \; {\underline{\underline{\sf{ \; Calculating \; the \; Perimeter \; :- }}}} [/tex]

[tex] \begin{gathered} \qquad \; {\green{:\implies}} \; \; \pink{\sf { Perimeter = \dfrac{Cost}{Rate} }} \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; :\implies \; \; \sf { Perimeter = \dfrac{75000}{100} } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; :\implies \; \; \sf { Perimeter = \cancel\dfrac{75000}{100} } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; :\implies \; \; {\underline{\boxed{\pmb{\orange{\frak{ Perimeter = 750 \; m }}}}}} \; \pmb\dag \\ \\ \\ \end{gathered} [/tex]

[tex] \\ \\ [/tex]

[tex] \maltese \; {\underline{\underline{\sf{ \; Let \; the \; Ratios \; :- }}}} [/tex]

• Length = 11y
• Breadth = 4y

[tex] \\ \\ [/tex]

[tex] \maltese \; {\underline{\underline{\sf{ \; Calculating \; the \; Value \; of \; y \; :- }}}} [/tex]

[tex] \begin{gathered} \qquad \; \longmapsto \; \; \sf { Perimeter = 2 \bigg( Length + Breadth \bigg) } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \longmapsto \; \; \sf { 750 = 2 \bigg( 11y + 4y \bigg) } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \longmapsto \; \; \sf { 750 = 2 \times 15y } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \longmapsto \; \; \sf { 750 = 30y } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \longmapsto \; \; \sf { \dfrac{750}{30} = y } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \longmapsto \; \; \sf { \cancel\dfrac{750}{30} = y } \\ \\ \\ \end{gathered} [/tex]

[tex] \begin{gathered} \qquad \; \longmapsto \; \; {\underline{\boxed{\pmb{\purple{\frak{ y = 25 }}}}}} \; \pmb\dag \\ \\ \\ \end{gathered} [/tex]

[tex] \\ \\ [/tex]

[tex] \maltese \; {\underline{\underline{\sf{ \; Calculating \; the \; Dimensions \; :- }}}} [/tex]

• Length = 11y = 11(25) = 275 m
• Breadth = 4y = 4(25) = 100 m

[tex] \\ \\ [/tex]

[tex] \therefore \; [/tex] Length of the plot is 275 m and the Breadth is 100 m .

[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]

:)