Question

3.
The length of a rectangle is twice its breadth If the perimeter is 72 m then find the length and
breadth of the rectangle.

Answer 1

Given :

• Length of the rectangle is twice its breadth
• Perimeter of the rectangle = 72 m

To find :

• Length of the rectangle
• Breadth of the rectangle

Concept :

Firstly, assume the breadth of the rectangle as any variable, let it be x. And the length of the rectangle will be twice of it,. i.e 2x. By using the formula of perimeter of the rectangle we will find the value of x. Substitute the value of x in the length and breadth of the rectangle which we've let, the resultant will be our required answer.

➳ Formula of perimeter of the rectangle :-

• Perimeter of the rectangle = 2(l + b)

where,

• l = length of the rectangle
• b = breadth of the rectangle

Solution :

Let,

• Breadth of the rectangle = x
• Length of the rectangle = 2x

Using formula,

★ Perimeter of the rectangle = 2(l + b)

Substituting the given values,

⇒ 72 = 2(2x + x)

Adding the values inside the brackets.

⇒ 72 = 2(3x)

Transposing 2 to the other side.

⇒ 72 ÷ 2 = 3x

⇒ 36 = 3x

Transposing 3 to the other side.

⇒ 36 ÷ 3 = x

⇒ 12 = x

The value of x = 12.

Substitute the value of x in th length and breadth of the rectangle.

Dimensions of the rectangle :-

• Length of the rectangle = 2x = 2 × 12 = 24 m
• Breadth of the rectangle = x = 12 m

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Verification :-

For verifying the value of length and breadth of the rectangle substitute the values in the formula of perimeter of the rectangle. If the resultant will be equal to 72 m as mentioned in the question, then the values are right.

⇒ Perimeter = 2(l + b)

⇒ Perimeter = 2(24 + 12)

⇒ Perimeter = 2(36)

⇒ Perimeter = 72

Perimeter of the rectangle = 72 m

Hence, verified.

Answer 2

Given : The length of a rectangle is twice its breadth & Perimeter of Rectangle is 72 m .

Exigency to find : The Length and Breadth of Rectangle.

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❍ Let's Consider the Breadth of Rectangle be x m

Given that ,

• The length of a rectangle is twice its breadth .

Therefore,

• Length of Rectangle is 2x m .

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{Perimeter _{(Rectangle)} \:: 2( l + b) }\bigg\rgroup$

⠀⠀⠀⠀⠀ Here l is the Length of Rectangle , b is the Breadth of Rectangle & we know that , Perimeter of Rectangle is 72 m

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \longmapsto \sf 2 ( x + 2x ) = 72$

$\qquad \longmapsto \sf x + 2x = \cancel {\dfrac{72}{2}}$

$\qquad \longmapsto \sf x + 2x = 36$

$\qquad \longmapsto \sf 3x = 36$

$\qquad \longmapsto \sf x = \cancel {\dfrac{36}{3}}$

$\qquad \longmapsto \frak{\underline{\purple{\:x = 12 \:m }} }\bigstar$

Therefore,

• Length of Rectangle is x = 12 m .
• Breadth of Rectangle is 2x = 2(12) = 24 m .

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:Length \:\&\:\:Breadth \:of\:Rectangle \:are\:\bf{12\:cm\:\:24\:\:cm}}}}$

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

$\dag\:\:\it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{Perimeter _{(Rectangle)} \:: 2( l + b) }\bigg\rgroup$

⠀⠀⠀⠀⠀ Here l is the Length of Rectangle , b is the Breadth of Rectangle & we know that , Perimeter of Rectangle is 72 m

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \longmapsto \sf 2 ( 12 + 24 ) = 72$

$\qquad \longmapsto \sf 2 ( 36 ) = 72$

$\qquad \longmapsto \frak{\underline{\purple{\:72m = 72 \:m }} }\bigstar$

⠀⠀⠀⠀⠀$\therefore {\underline {\bf{ Hence, \:Verified \:}}}$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

$\qquad \leadsto \sf Area_{(Rectangle)} = Length \times Breadth$

$\qquad \leadsto \sf Perimeter _{(Rectangle)} = 2 (Length + Breadth)$

$\qquad \leadsto \sf Area_{(Square)} = Side \times Side$

$\qquad \leadsto \sf Perimeter _{(Square)} = 4 \times Side$

$\qquad \leadsto \sf Area_{(Trapezium)} = \dfrac{1}{2} \times Height \times (a + b )$

$\qquad \leadsto \sf Area_{(Parallelogram)} = Base \times Height$

$\qquad \leadsto \sf Area_{(Triangle)} = \dfrac{1}{2} \times Base \times Height$

$\qquad \leadsto \sf Area_{(Rhombus)} = \dfrac{1}{2} \times Diagonal _{1}\times Diagonal_{2}$

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