Question

A Rectangular plot has length of 25 metre and breadth 15 metre. How much fencing will be required to fence the plot?

Answer 1

$\huge{\boxed{\mathtt{\purple{Answer}}}}$

Breadth of rectangle 50m
Let length be l
The cost of of fencing 18 per metre
The total cost 4680
Perimeter of Rectangular plot

18
4680
​
=260
The perimeter is given by 2(l+b)=260
2(l+50)=260
l+50=130
l=80
The length of rectangular plot is 80m
The area of plot is given by lb=(80)(50)
4000m ^2

Cost of leveling for 1m ^2
is 7.6m ^2

Cot of lrvrling for 4000m^2
=4000×7.6=30400Rs.
The cost of leveling is 30400Rs.

Answer 2

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\bold{ \underline{ \blue{ Given, }}} \\ \\ \texttt{ Length of the rectangular plot, l = 25 m} \\ \texttt{Breadth of the rectangular plot, b = 15 m }$

$\bold {\underline{ \blue{ \: To \: \: \: \: find :- } \: }} \\ \\ \: \: \: \: \: \: \: \: \sf{How \: \: many \: \: metre \: \: of \: \: fencing \: \: is \: \: </p><p>required \: \: to \: \: enclose \: \: } \\ \: \: \: \: \sf{the \: \: rectangular \: \: plot. }$

$\underline{ \sf{ \blue{ \: \: We \: \: know \: \: that, \: \: }}} \\ \\ \boxed{ \pink{ \bold{Perimeter \: \: of \: \: a \: \: rectangle = 2 \times (Length + Breadth)}}}$

$\bold{ \underline{ \blue{ \: According \: \: to \: \: question, \: }}}$

$\sf{Perimeter \: \: of \: \: the \: \: rectangular \: \: plot = 2 \times (l + b)} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{= \{2 \times (25 + 15 ) \} \: \: m } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = (2 \times 40) \: \: m} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = 80 \: \: m }$

$\bold{ \therefore \: Perimeter \: \: of \: \: the \: \: rectangular \: \: plot = 80 \: \: m}$

$\: \: \: \: \: \: \: \: \: \texttt{Hence, \red{ \underline{80 metres}} of fencing is</p><p>required to } \\ \texttt{enclose a rectangular plot of length 25 m and } \\ \texttt{breadth 15 m.}$