Question

Challenge!
1. A rectangular field measure 70 m by 50 m. Each side is to be fenced with four
rows of wires. Find the length of the wire needed​

Answer 1

Answer :-

Given :-

• Length = 70 m
• Breadth = 50 m

To Find :-

• Length of wire required to fence four rows of wire.

Solution :-

Here, rectangular field is to be fenced with wires. So,

Length of wire required = Perimeter of rectangular field

Perimeter of rectangular field = 2 ( l + b )

= 2 ( 50 + 70 )

= 2 ( 120 )

= 240 m

Hence, The length of wire needed = 240 m.

Additional information -

Rectangle -

• Area of rectangle = Length × Breadth

Square -

• Perimerer of square = 4 × side
• Area of square = (side)²

Triangle -

• Perimeter of triangle = Sum of sides
• Area of triangle = ½ × Base × Height

Rhombus -

• Perimeter of rhombus = 4 × side
• Area of rhombus = ½ × Product of diagnols

Parallelogram -

• Perimeter = 2 × ( sum of adjacent sides )
• Area of parallelogram = Base × Height

Answer 2

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✪ Challenge Accepted !!! ツ

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$\LARGE\mathfrak{\underline{\underline\textcolor{aqua}{Given :-}}}$

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$\qquad\tt{:}\longrightarrow\large\textsf{Length of the rectangular field = 70m }\\\\\\\qquad\tt{:}\longrightarrow\large\textsf{ Breadth of the rectangular field = 50m }$

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$\LARGE\mathfrak{\underline{\underline\textcolor{aqua}{To \; \; Find :-}}}$

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• Length of the wire required to fence the rectangular field four times = ?

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$\LARGE\mathfrak{\underline{\underline\textcolor{aqua}{How \; \; To \; \; Solve :-}}}$

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• Now first you have to know that fencing any region or any field means to put wires around the perimeter of the rectangular field.

• So first we need to find the perimeter of the rectangular field and then multiply it by four.

• By doing this we can get the required answer for the given Question.

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$\LARGE\mathfrak{\underline{\underline\textcolor{aqua}{Formula :-}}}$

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$\boxed{\large\textsf{ {\large\textsf\textcolor{purple}{Perimeter}}_{\large\textsf\textcolor{purple}{( \; Rectangle \; )}} \large\textsf\textcolor{purple}{= 2 ( l + b )} }}$

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• Length ( l ) = 70m
• Breadth ( b ) = 50m

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$\LARGE\mathfrak{\underline{\underline\textcolor{aqua}{Solution :-}}}$

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$\large\textsf{ {\large\textsf\textcolor{orange}{Perimeter}}_{\large\textsf\textcolor{orange}{( \; Rectangle \; )}}\large\textsf\textcolor{orange}{= 2 ( l + b )} }\large: \: \Longrightarrow\textsf{ \; \; = 2 ( 70 + 50 )}\\\\\\\\\large: \: \Longrightarrow\textsf{\; \; = 2 × 120}\\\\\\\\\large: \: \Longrightarrow\textsf{\; \; = 240m}\\\\\\\\\boxed{\large\textsf{ {\large\textsf\textcolor{red}{Perimeter}}_{\large\textsf\textcolor{red}{( \; Rectangle \; )}} \large\textsf\textcolor{red}{= 240m} }}$

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• As we know the Perimeter of the Field , so we need to multiply this value by 4 so that we can get the total length of the wire required to fence the rectangular field four times .

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$\qquad\tt{:}\longrightarrow\large\textsf{\; \; 4 × Perimeter of the Field}$

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$\qquad\tt{:}\longrightarrow\large\textsf{\; \; 4 × 240}$

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$\qquad\tt{:}\longrightarrow\boxed{\underline{\overline{\large\textsf\textcolor{red}{\; \; 960m}}}}$

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∴ The total length of the wire required to fence the rectangular field = 960m