Question

The perimeter of a classroom is 42m. If the length of the classroom is 12m,
find the breadth of the classroom and hence find the area of the classroom.
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Answer 1

G I V E N :

• The perimeter of a classroom is 42m. If the length of the classroom is 12m.

U N K N O W N :

• Breadth of classroom = ?
• Area of classroom = ?

S O L U T I O N :

• In this question the shape of classroom is not given so we will assume the shape of classroom as rectangle.
• Perimeter of classroom = 42 m
• Length of classroom = 12 m

First of all we will calculate the Breadth of classroom :

Let Breadth of rectangular classroom be 'b'.

→ Perimeter of rectangle = 2(Length + Breadth)

→ 42 = 2(12 + b)

→ 42 = 24 + 2b

→ 42 - 24 = 2b

→ 18 = 2b

→ 2b = 18

→ b = 18 ÷ 2

→ b = 9 m

Hence, the breadth of rectangular classroom is 9 m.

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V E R I F I C A T I O N :

➺ 42 = 2(12 + b)

➺ 42 = 2(12 + 9)

➺ 42 = 2(21)

➺ 42 = 42

H E N C E V E R I F I E D :

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Now, Let's find the area of rectangular classroom :

➻ Area of rectangle = Length × Breadth

➻ Area of rectangle = 12 × 9

➻ Area of rectangle = 108 m²

Hence, the area of rectangular classroom is 108 m² .

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Answer 2

$\large{\boxed{\boxed{\sf Let's \: Understand \: Question \: F1^{st}}}}$

It is an simple question in which we have given Perimeter and Length of a classroom. and have to find its Breadth and Area.

$\large{\boxed{\boxed{\sf How \: To \: Do \: this?}}}$

Here, f1st we apply formula of Perimeter of rectangle and by substituting given values in it we will get it's breadth as our f1st answer. Then we will apply formula of area of rectangle substitute values in this one we will got our 2nd answer.

$\huge{\boxed{\underline{\sf AnSwer}}}$

_____________________________

Given:-

✻ Perimeter of classroom = 42m

✻ Length of classroom = 12m

Find:-

❅ Breadth of classroom.

❅ Area of classroom.

Formula Used:-

$\huge{\boxed{\sf Perimeter \: of \: rectangle = 2(l + b)}}$

$\huge{\boxed{\sf Area \: of \: rectangle = l \times b}}$

Solution:-

Let, Breadth of Rectangle be 'b' m

Now, using

$\underline{\boxed{\sf Perimeter \: of \: rectangle = 2(l + b)}}$

$\sf where\small{\begin{cases} \sf Perimeter = 42m \\ \sf Length, l = 12m\end{cases}}$

$\pink\bigstar$ Substituting these values:-

$\implies\sf Perimeter = 2(l + b)$

$\implies\sf 42 = 2(12 + b)$

$\implies\sf 42 = 24 +2b$

$\implies\sf 42 - 24=2b$

$\implies\sf 18=2b$

$\implies\sf \dfrac{18}{2}=b$

$\implies\sf 9m=b$

$\huge{\underline{\boxed{ \sf\therefore Breadth\:of\: classroom\:is\:9m}}}$

Now, using

$\underline{\boxed{\sf Area \: of \: rectangle = l \times b}}$

$\sf where\small{\begin{cases} \sf Length, l = 12m \\ \sf Breadth, b = 9m \end{cases}}$

$\red\bigstar$ Substituting these values:-

$\dashrightarrow\sf Area= l \times b$

$\dashrightarrow\sf Area=12 \times 9$

$\dashrightarrow\sf Area=108 {m}^{2}$

$\huge{\underline{\boxed{ \sf\therefore Area\:of\: classroom\:is\:108 {m}^{2} }}}$