Question

A rectangular field is 450 m long and 318 m wide. It is to be fenced with a barbed
wire. How much wire will be required?
​

Answer 1

Answer:

1086 m wire

450 m +318m ×2

=1086 m wire required

Answer 2

★ Given :

• Length of the field is 450 m
• Breadth of the field is 318 m

★ To Find :

• How much wire will be required?

★ Solution :

• We know that

$\star\underline{\boxed{\sf{ \pink{ Perimeter _{(rectangle)} = 2(Length \: +\: Breadth)}}}}$

• Now, Substitute the values :-

$\implies{\rm{ Perimeter = 2(450 + 318)}}$

$\implies{\rm{ Perimeter = 2(768)}}$

$\implies{ \boxed{ \purple{\rm{ Perimeter =1536 \: m}}}}$

${ \underline{ \rm{ Diagram :-}}}$

$\begin{gathered}\begin{gathered} \pink{\tt450m} \: \: \: \: \: \: \: \: \: \: \: \\ \pink{\boxed{\begin{array}{}\bf { \red{}}\\{\qquad \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }{}\\ { \sf{ }}\\ { \sf{ }} \\ \\ { \sf{ }}\end{array}}} \pink{ \tt \:318m} \end{gathered}\end{gathered}$

${ \underline{ \rm{ More \: Information :-}}}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\begin {minipage}{9cm}\\ \dag\quad \small\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {minipage}}\end{gathered}\end{gathered}\end{gathered}\end{gathered < u > }\end{gathered}\end{gathered}$