Question

The areas of a rectangle and a square are equal. If the length of the rectangle is 16 cm and
breadth is 9 cm, find the side of the square.​

Answer 1

Answer:

12 cm

Explanation:

Area of rectangle = Area of square

l × b = a sq

16 × 9 = a sq

144 = a sq

a = √144

a = 12

Side of square = 12 cm

Answer 2

Question:

The areas of a rectangle and a square are equal. If the length of the rectangle is 16 cm and breadth is 9 cm, find the side of the square.

To find:

• side of square

Given:

• Area of rectangle = Area of square
• Length of rectangle= 16cm
• Breadth of rectangle = 9cm

Formula used:

$\begin{gathered}\\\;\sf{\to\;\;Area\;of\;Square\;=\;(Side)^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\to\;\;Area\;of\;Rectangle\;=\;Length\;\times\;Breadth}\end{gathered}$

Answer:

First Let's find area of rectangle:

$\bigstar \boxed{ \rm Area \: of \: reactangle = length \times breadth}$

by using this formula we can find value of Area of rectangle

$: \implies \sf Area \: of \: reactangle = length \times breadth$

put values of length and breadth

$: \implies \sf Area \: of \: reactangle = 16 \times 9$

$: \implies \underline{ \underline{ \sf Area \: of \: reactangle =144 {cm}^{2}}}$

Now as we know Area of rectangle = Area of square

.•. Area of square =144cm²

Now as we know:

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

by using this formula we can find value of its sides

$: \implies \sf Area \: of \: square = {side}^{2}$

put value of Area of square

$: \implies \sf 144 = {side}^{2}$

$: \implies \sf \sqrt{144} = side$

$: \implies \sf \sqrt{12 \times 12} = side$

$: \implies \underline{ \underline{ \sf side = 12 \: cm}}$

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☆Know more ☆

$\begin{gathered}\\\;\sf{\longrightarrow \;\;Perimeter\;of\;Square\;=\;4\;\times\;Side}\end{gathered}$

$\begin{gathered}\\\;\sf{\longrightarrow \;\;Perimeter\;of\;Rectangle\;=\;2\;\times\;(Length\;+\;Breadth)}\end{gathered}$

$\begin{gathered}\\\;\sf{\longrightarrow\;\;Area\;of\;Triangle\;=\;\dfrac{1}{2}\;\times\;Base\;\times\;Height}\end{gathered}$

$\begin{gathered}\\\;\sf{\longrightarrow \;\;Area\;of\;Circle\;=\;\pi r^{2}}\end{gathered}$

$\begin{gathered}\\\;\sf{\longrightarrow \;\;Area\;of\;Parallelogram\;=\;Base\;\times\;Height}\end{gathered}$

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And all we are done! ✔

:D