Math, asked by amitbhattal976, 2 months ago

The area of a square field with side 45 m is the same as the area of rectangular field with breadth 35 m. Find the length and perimeter of the rectangular field​

Answers

Answered by DüllStâr
77

\Large\pink {\underline {\mathfrak{~~~~~~~~~~~~~\purple {Required\:Solution}~~~~~~~~~~~~~}}}

To find:

  • Perimeter of rectangle

  • Length of rectangle

Given:

  • Side of square= 45 m

  • Area of square = Area of rectangle

  • Breadth of rectangle = 35 m

Solution:

We know:

\bigstar \boxed {\rm{}Area~of~square=side^{2} }

 \\

By using these Formula we can find area of square

 \\

 \dashrightarrow\sf{}Area~of~square=side^{2}  \\

 \\

 \dashrightarrow\sf{}Area~of~square=(45)^{2}  \\

 \\

 \dashrightarrow\sf{}Area~of~square = 45 \times 45 \\

 \\

 \dashrightarrow  \underline{\boxed{\sf{}Area~of~square = 2025 {m}^{2} }} \\

 \\

Now we know:

 \\

 \dashrightarrow\sf{}Area~of~square=Area~of~rectangle \\

 \\

 \therefore\sf{}Area~of~rectangle = 2025 {m}^{2}  \\

 \\

We know:

\bigstar \boxed {\rm{}Area~of~rectangle=length \times breadth}

By using this formula we can find value of length of Rectangle

 \dashrightarrow\sf{}Area~of~rectangle=length \times breadth \\

 \\

 \dashrightarrow\sf{}2025=length \times 35 \\

 \\

 \dashrightarrow\sf{} \dfrac{2025}{35} =length  \\

 \\

 \dashrightarrow\sf{} length = \dfrac{2025}{35}  \\

 \\

 \dashrightarrow\sf{} length = \dfrac{\cancel{2025}}{\cancel{35}}  \\

 \\

 \dashrightarrow\sf{} length = \dfrac{405}{7}  \\

 \\

 \dashrightarrow \underline{ \boxed{\sf{} length = 57.9m}} \\

 \\

 \therefore\underline{\sf{} length \: of \: rectangle = 57.9m}\\

 \\

Now Let's find perimeter:

 \\

We know:

\bigstar \boxed {\rm{}perimeter~of~rectangle=2(length + breadth)}

 \\

\dashrightarrow\sf{}perimeter~of~rectangle=2(length + breadth) \\

 \\

\dashrightarrow\sf{}perimeter~of~rectangle=2(57.9+35) \\

 \\

\dashrightarrow\sf{}perimeter~of~rectangle=2(57.9)+2(35) \\

 \\

\dashrightarrow\sf{}perimeter~of~rectangle=115.8+70 \\

 \\

\dashrightarrow  \underline{\boxed{\sf{} perimeter~of~rectangle=185.8 \: m}} \\

 \\

\therefore \underline{\sf{} perimeter~of~rectangle=185.8 \: m}\\

Answered by XxMissCutiepiexX
7

Step-by-step explanation:

\Large\pink {\underline {\mathfrak{~~~~~~~~~~~~~\purple {Required\:Solution}~~~~~~~~~~~~~}}}

To find:

  • Perimeter of rectangle

  • Length of rectangle

Given:

  • Side of square= 45 m

  • Area of square = Area of rectangle

  • Breadth of rectangle = 35 m

Solution:

  • We know:

\bigstar \boxed {\rm{}Area~of~square=side^{2} }

 \\

By using these Formula we can find area of square

 \\

 \dashrightarrow\sf{}Area~of~square=side^{2}  \\

 \\

 \dashrightarrow\sf{}Area~of~square=(45)^{2}  \\

 \\

 \dashrightarrow\sf{}Area~of~square = 45 \times 45 \\

 \\

 \dashrightarrow  \underline{\boxed{\sf{}Area~of~square = 2025 {m}^{2} }} \\

 \\

Now we know:

 \\

 \dashrightarrow\sf{}Area~of~square=Area~of~rectangle \\

 \\

 \therefore\sf{}Area~of~rectangle = 2025 {m}^{2}  \\

 \\

We know:

\bigstar \boxed {\rm{}Area~of~rectangle=length \times breadth}

By using this formula we can find value of length of Rectangle

 \dashrightarrow\sf{}Area~of~rectangle=length \times breadth \\

 \\

 \dashrightarrow\sf{}2025=length \times 35 \\

 \\

 \dashrightarrow\sf{} \dfrac{2025}{35} =length  \\

 \\

 \dashrightarrow\sf{} length = \dfrac{2025}{35}  \\

 \\

 \dashrightarrow\sf{} length = \dfrac{\cancel{2025}}{\cancel{35}}  \\

 \\

 \dashrightarrow\sf{} length = \dfrac{405}{7}  \\

 \\

 \dashrightarrow \underline{ \boxed{\sf{} length = 57.9m}} \\

 \\

 \therefore\underline{\sf{} length \: of \: rectangle = 57.9m}\\

 \\

Now Let's find perimeter:

 \\

We know:

\bigstar \boxed {\rm{}perimeter~of~rectangle=2(length + breadth)}

 \\

\dashrightarrow\sf{}perimeter~of~rectangle=2(length + breadth) \\

 \\

\dashrightarrow\sf{}perimeter~of~rectangle=2(57.9+35) \\

 \\

\dashrightarrow\sf{}perimeter~of~rectangle=2(57.9)+2(35) \\

 \\

\dashrightarrow\sf{}perimeter~of~rectangle=115.8+70 \\

 \\

\dashrightarrow  \underline{\boxed{\sf{} perimeter~of~rectangle=185.8 \: m}} \\

 \\

\therefore \underline{\sf{} perimeter~of~rectangle=185.8 \: m}\\

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