Question

10. If the length of a rectangle is doubled and breadth is halved, then what would be the effect on its
perimeter and area?​

Answer 1

Solution :

Let us assume that the rectangle initially has a length of a units and a breadth of b units .

Area Of Rectangle :

=> [Length × Breadth ]

=> ab

Perimeter of the rectangle :

=> 2 [Length + Breadth ]

=> 2[ a + b]

Now , the length of the rectangle is doubled and the breadth is halved.

New Length :

=> 2a

New Breath :

=> b/2

New Area :

=>{ 2a } × { b/2}

=> ab

New Perimeter ;

=> 2 [ 2a + b/2 ]

=> 2 [ 4 a + b]/2

=> 4a + b

Thus , we can observe that the area remains constant while the perimeter changes by 2a - b units .

This is the required answer .

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Answer 2

Lᴇᴛ,

• Length of a rectangle is x unit.

• Breadth of the rectangle is y unit.

Wᴇ ᴋɴᴏᴡ ᴛʜᴀᴛ,

$\red\bigstar\:\:{\underline{\purple{\boxed{\bf{\green{Perimeter_{(rec.)}\:=\:2\:(Length\:+\:Breadth)}}}}}}$

$:\longrightarrow\:\:\bf{Perimeter\:=\:2\:(x\:+\:y)}--(1)$

Aɢᴀɪɴ,

$\pink\bigstar\:\:{\underline{\orange{\boxed{\bf{\purple{Area_{(rec.)}\:=\:Length\times{Breadth}}}}}}}$

$\longmapsto\:\:\bf{Area\:=\:x\times{y}}$

$\longmapsto\:\:\bf{Area\:=\:x\:y}--(2)$

Aᴄᴄᴏʀᴅɪɴɢ ᴛᴏ ᴛʜᴇ ǫᴜᴇsᴛɪᴏɴ,

• Length of the rectangle is doubled, i.e. 2x.

• Breadth is halved, i.e. $\bf{\dfrac{y}{2}}$.

$:\longrightarrow\:\:\bf{Perimeter\:=\:2\:\Big(2x\:+\:\dfrac{y}{2}\Big)}$

$:\longrightarrow\:\:\bf{Perimeter\:=\:2\times{\dfrac{4x\:+\:y}{2}}}$

$:\longrightarrow\:\:\bf{Perimeter\:=\:4x\:+\:y}--(3)$

Sᴏ,

✒ Change in perimeter is,

➛ equation (3) - equation (1)

➛ (4x + y) - [2 × (x + y)]

➛ 4x + y - (2x + 2y)

➛ 4x + y - 2x - 2y

➛ 2x - y

$\Large\bf\orange{Therefore,}$

Perimeter of the rectangle is changes by (2x - y) units.

Aɴᴅ,

$\longmapsto\:\:\bf{Area\:=\:2x\times{\dfrac{y}{2}}}$

$\longmapsto\:\:\bf{Area\:=\:x\:y}--(4)$

Sᴏ,

✒ Change in area is,

↣ equation (4) - equation (2)

↣ xy - xy

↣ 0

$\Large\bf\green{Therefore,}$

Area of the rectangle is same as before (no change).