Question

A rectangular piece of paper has a length of 14 cm and a breadth of 12 cm. A square piece of paper of perimeter 24 cm is cut off from it. Find the area of the piece of paper left.
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Answer 1

☘️ ☘️ ☘️ Heya mate!!! here is your awesome answer... ☘️ ☘️ ☘️

Length of rectangular piece of paper = 14cm

Breadth of rectangular piece of paper = 12cm

perimeter of paper cut from it = 24

side of the square piece cut from it = 24 / 4 = 6

area of square piece cut from it = 6 * 6 = 36

area of the rectangular piece = 14 * 12 = 168

area left = area of main piece - area of square peice

= 168 - 36

= 132$cm^{2}$ is the area of paper piece left

❄️ have an awesome day mate!!! ❄️

Answer 2

$\: \: \: \: \:{\large{\pmb{\sf{\underline{ Here's \: your \: required \: solution!! }}}}}$

• Here, we asked to find the area of the remaining piece of paper. As we are given length of paper is 14 cm and breadth of paper is 12 cm. Let's it's area! (Consider that it's a rectangle)

$\\ :\implies{ \textsf { \textsf\purple{{Area = Length × Breadth \: sq.cm}}}} \\ \\ :\implies { \sf{Area = (14 \times 12)}} \\ \\ :\implies{ \boxed{ \sf{Area = \: }{ \frak{ \purple{168\: {cm}^{2}}}}}}$

Again, perimeter of the square is 24cm.Generally, it's given to find out the side of the square. So, let's find!

• Let us assume that the side of the given square be "a".

$\\ :\implies{ \textsf { \textsf\green{{Perimeter = 4 × a \: cm}}}} \\ \\ :\implies { \sf{24 = (4 \times a)}}\\ \\ :\implies{ \sf{ \dfrac{ \cancel{24}}{ \cancel{4} }= a}} \\ \\ :\implies{ \boxed{ \sf{Side = \: }{ \frak{ \green{6\: cm}}}}}$

Then, the area of the square will be :-

$:\implies{\sf{\purple{Area_{(Square )}=(Side)^2\:sq.cm}}}$

$\:\:\:\:\:\:\:\::\implies{\sf{6\times 6}}$

$\:\:\:\:\:\:\:\::\implies\purple{{\sf{36}}}$

Area of remaining Piece of paper :-

$\small\leadsto{\sf{\green{Area\:of\:Main\:Piece-Area\:of\:square\:piece}}}$

$\\ \:\:\:\:\:\::\implies{\sf{168-36}}$

$\\ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\::\implies{\boxed{\frak{\green{132\:cm^2}}}}$

$\:$

$\\\therefore\underline{\sf{The\:area\:of\:the\: Main\:paper \: piece \:is\:168\:cm^2\: whereas,\:the\:area\:of\: the\: paper\: piece \:left \:132\:cm^2.}}$

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