Question

If the side of a square is doubled, how many times the perimeter of the first square will that of the new square be?​

Answer 1

${\blue{\underline{\underline{\purple{\textsf{\textbf{Answer : }}}}}}}$

➙The perimeter of the new square will be 2 times the original square

${\blue{\underline{\underline{\purple{\textsf{\textbf{Step-by-step explanation: }}}}}}}$

Question says that, if the side of a square is doubled find that how many times the perimeter of the square will be that of the new square be.

________________________________

Step 1 : Finding the perimeter of the square :

Let's assume, the side of the square as $x \: \sf units$

Perimeter of a square is given by :

→ 4 × each side

→ 4 × $x$

→ $4x \: \sf units$

________________________________

Step 2 : Finding the perimeter after doubling the each side of the square :

The side of the square doubles

So, the each side now will be :

$\to 2x$

Perimeter of the new square :

$\to 2x \times 4$

$\to 8x \sf units.$

________________________________

Step 3 : On comparing the perimeters :

The perimeter increased :

$\to \dfrac{p_2}{p_1}$

• $p_1$ is the perimeter of the original square.
• $p_2$ is the perimeter of the new square.

So,

$\to \dfrac{8x}{4x}$

$\to 2 \sf units$

${ \underline{ \sf{ \therefore{The \: perimeter \:is \: 2 \: times \: the \: perimeter \: of \: the \: first \: square}}}}$

____________________

Answer 2

$\huge \mathfrak{ \blue{An}} \mathfrak{ \purple{sw}} \mathfrak{ \pink{er}}$

The perimeter of the new square will be 2 times the original square.

$\large \mathfrak{ \orange{Explanation : }}$

${ \green{\textsf {\textbf{Step 1 : }}}}$

Finding the perimeter of the square :

Let's assume, the side of the square as $x \: \sf units$

Perimeter of a square is given by :

→ 4 × each side

→ 4 × xx

→ $4x \: \sf units$

_____________________________

${ \green{ \textsf{ \textbf{Step \: 2  : }}}}$

Finding the perimeter after doubling the each side of the square :

The side of the square doubles.

So, the each side now will be :

→2x

Perimeter of the new square :

→2x × 4

$\to 8x \sf units$

_____________________________

${\green{\textsf{\textbf{Step 3 : }}}}$

On comparing the perimeters :

The perimeter increased :

$\to \dfrac{p_2}{p_1}$

• p_1p1 is the perimeter of the original square.
• p_2p2 is the perimeter of the new square.

So,

$\to \dfrac{8x}{4x}$

$\to 2 \sf units$

${\underline{ \sf{\red{ \therefore{The \: perimeter \:is \: 2 \: times \: the \: perimeter \: of \: the \: first \: square}}}}}$

_____________________________