Question

A rectangular sheet of paper with length 20cm and breadth 10cm is to be cut
into small squares of side 5cm each.
a. Find the area of one square piece?
b. Find the area of the rectangular sheet?
c. How many square pieces will be obtained

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Answer 1

QUESTION:

A rectangular sheet of paper with length 20cm and breadth 10cm is to be cut

into small squares of side 5cm each.

a. Find the area of one square piece?

b. Find the area of the rectangular sheet?

c. How many square pieces will be obtained

ANSWER:

Area of rectangular sheet of paper: L=20 cm ; B=10 cm

Area=l*b

20*10 = 200 cm square

Area of one small square: a = 5 cm

Area = a*a

5*5 = 25 cm square

number square pieces will be obtained = 200/ 25

= 8 number of square pieces

Answer 2

Step-by-step explanation:

Here the concept of Areas of Square and Areas of Rectangle has been used. We are already given dimensions of square and rectangle. So we can easily find out their Areas. Also, we know that sum of areas of every square piece will be equal to Area of Rectangular sheet. So, number of pieces will be the ratio of area of rectangular sheet by area of each square piece.

Let's do it !!

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★ Equations Used :-

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

AreaofSquare=(Side)

2

\begin{gathered}\\\;\boxed{\sf{Area\;of\;Rectangle\;=\;\bf{Length\;\times\;Breadth}}}\end{gathered}

AreaofRectangle=Length×Breadth

\begin{gathered}\\\;\boxed{\sf{Number\;of\;Pieces\;=\;\bf{\dfrac{Area\;of\;Rectangular\;Sheet}{Area\;of\;each\;Square\;piece}}}}\end{gathered}

NumberofPieces=

AreaofeachSquarepiece

AreaofRectangularSheet

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★ Solution :-

Given,

» Length of Rectangular Piece = 20 cm

» Breadth of Rectangular Piece = 10 cm

» Side of each Square Piece = 5 cm

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a.) For the Area of each Square Piece :-

\begin{gathered}\\\;\;\;\sf{:arrow\;\;Area\;of\;Square_{(each\;piece)}\;=\;\bf{(Side)^{2}}}\end{gathered}

:arrowAreaofSquare

(eachpiece)

=(Side)

2

\begin{gathered}\\\;\;\;\sf{:arrow\;\;Area\;of\;Square_{(each\;piece)}\;=\;\bf{(5)^{2}}}\end{gathered}

:arrowAreaofSquare

(eachpiece)

=(5)

2

\begin{gathered}\\\;\;\;\bf{:arrow\;\;Area\;of\;Square_{(each\;piece)}\;=\;\bf{25\;\;cm^{2}}}\end{gathered}

:arrowAreaofSquare

(eachpiece)

=25cm

2

\begin{gathered}\\\;\underline{\boxed{\tt{Area\;\;of\;\;Each\;\;Square\;\;Piece\;\;is\;\;\bf{25\;\;cm^{2}}}}}\end{gathered}

AreaofEachSquarePieceis25cm

2

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b.) For the Area of Rectangular Sheet :-

\begin{gathered}\\\;\;\;\sf{:\Longrightarrow\;\;Area\;of\;Rectangle\;=\;\bf{Length\;\times\;Breadth}}\end{gathered}

:⟹AreaofRectangle=Length×Breadth

\begin{gathered}\\\;\;\;\sf{:\Longrightarrow\;\;Area\;of\;Rectangle\;=\;\bf{20\;\times\;10}}\end{gathered}

:⟹AreaofRectangle=20×10

\begin{gathered}\\\;\;\;\bf{:\Longrightarrow\;\;Area\;of\;Rectangle\;=\;\bf{200\;\;cm^{2}}}\end{gathered}

:⟹AreaofRectangle=200cm

2

\begin{gathered}\\\;\underline{\boxed{\tt{Area\;\;of\;\;Rectangular\;\;Sheet\;\;is\;\;\bf{200\;\;cm^{2}}}}}\end{gathered}

AreaofRectangularSheetis200cm

2

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c.) For the Number of Square Pieces :-

\begin{gathered}\\\;\;\;\sf{:\mapsto\;\;Number\;of\;Pieces\;=\;\bf{\dfrac{Area\;of\;Rectangular\;Sheet}{Area\;of\;each\;Square\;piece}}}\end{gathered}

:↦NumberofPieces=

AreaofeachSquarepiece

AreaofRectangularSheet

\begin{gathered}\\\;\;\;\sf{:\mapsto\;\;Number\;of\;Pieces\;=\;\bf{\dfrac{200}{25}}}\end{gathered}

:↦NumberofPieces=

25

200

\begin{gathered}\\\;\;\;\sf{:\mapsto\;\;Number\;of\;Pieces\;=\;\bf{8}}\end{gathered}

:↦NumberofPieces=8

\begin{gathered}\\\;\underline{\boxed{\tt{Number\;\;of\;\;Square\;\;Pieces\;\;is\;\;\bf{8\;\;pieces}}}}\end{gathered}

NumberofSquarePiecesis8pieces

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★ More to know :-

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

⇝AreaofCircle=πr

2

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

⇝AreaofTriangle=

2

1

×Base×Height

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

⇝AreaofParallelogram=Base×Height

\begin{gathered}\\\;\sf{\leadsto\;\;Perimeter\;of\;Circle\;=\;2\pi r}\end{gathered}

⇝PerimeterofCircle=2πr

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

⇝PerimeterofRectangle=2×(Length+Breadth)

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

⇝PerimeterofSquare=4×Side

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