Question

) A picture is to be painted on a canvas of length 40 cm and width 80 cm. A margin of 5 cm has to be left on all sides of the canvas. Find the area of the actual picture and the area of the margin in sq m

Answer 1

$\sf\underbrace{Answer:-}$

• Area of actual picture = 0.21 m².
• Area of margin= 0.11 m².

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$\color{pink}{\textbf{\textsf{Explaination:-}}}$

Given,

• Width and length of the canvas is 40 cm and 80 cm respectively.
• Length of the margin 5 cm. (along the canvas).

$\sf\Large{\underline{\underline{To \: Find:-}}}$

• Area of Actual picture, and area of the margin in sq. m.

$\sf\Large{\underline{\underline{Formula \:to\: be \: used:-}}}$

$\\underline{\boxed{\sf{{area}_{(Rectangle)}=(l \times b)\:units}}}$

• l = length
• b = breadth

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$\sf\Large{\underline{\underline{Solution:-}}}$

We are said that a picture is painted on a canvas whose length = 40 cm, and breadth = 80 cm. And there is a margin of 5 cm from all sides along the canvas.

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We've to find the area of actual picture (excluding Margin) and the area occupied by margin in square m.

So, let's find the area of whole canvas first by inserting the given values in the Formula of Area of Rectangle.

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area_{(rectangle)}\;=\;\bf{{l \times b}\:}}\end{gathered}$

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area_{(rectangle)}\;=\;\bf{{40 \times 80}\:}}\end{gathered}$

$\: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area_{(rectangle)}\;=\;\bf{{3200} {cm}^{2} \:}}\end{gathered}$

Thus, the area of canvas (ABCD) is 3200 cm².

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• Finding PQRS.

According to question, the canvas has a margin of 5 cm along it's side. So, subtract 5 cm from all four sides of ABCD, to get the dimensions of PQRS. Find area in sq. m. (Refer to the attachment for better understanding).

RS = length = 40cm - (5cm + 5cm) = 30cm

QS = Breadth = 80cm - (5cm + 5cm) = 70 cm

Put the obtained values in the Formula,

$\begin{gathered}\\\;\sf{:\rightarrow\;\;area_{(PQRS)}\;=\;\bf{{30 \times 70}\:}}\end{gathered}$

$\: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area_{(PQRS)}\;=\;\bf{{2100 {cm}^{2} }\:}}\end{gathered}$

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• Converting the area in sq. m.

$\begin{gathered}\\\;\sf{\;\;we \: know \: that \: 1 {cm}^{2} \;=\;\bf{ \frac{1}{10000} {m}^{2} \:}}\end{gathered}$

$\begin{gathered}\\\;\sf{\;\;so \: 2100 {cm}^{2}\;=\;\bf\bigg({ \frac{1}{10000} \times 2100\bigg){m}^{2} \:}}\end{gathered}[\tex]</p><p></p><p>[tex]\begin{gathered}\:{\sf{\dfrac{{1}}{\cancel{10000}} \times 21{\cancel{00}}}}\end{gathered}[\tex]</p><p>[tex]\begin{gathered}\\\;\sf{:\rightarrow\;\;area_{(in \: sq \: m)}\;=\;\bf{{0.21 {m}^{2} }\:}}\end{gathered}[\tex]</p><p></p><p>[tex]{\underline{\underline{\sf{\green{ Area \: of \: PQRS \: (Actual \: picture) \: is \: 0.21m². }}}}}$

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Now, To Find the area of margin, subract PQRS from ABCD.

$\: \: \: \: \: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area_{(margin)}\;=\;\bf{{3200} {cm}^{2} - 2100 {cm}^{2} \:}}\end{gathered}$

$\: \: \: \: \begin{gathered}\\\;\sf{:\rightarrow\;\;area_{(margin)}\;=\;\bf{1100 {cm}^{2} \:}}\end{gathered}$

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• by converting it in sq. m, we get

$\begin{gathered}\\\;\sf{\;\;so \: 1100 {cm}^{2}\;=\;\bf\bigg({\frac{1}{10000} \times 1100\bigg){m}^{2} \:}}\end{gathered}[\tex]</p><p></p><p>[tex]\begin{gathered}\:{\sf{\dfrac{{1}}{\cancel{10000}} \times 11{\cancel{00}}}}\end{gathered}[\tex]</p><p>[tex]\begin{gathered}\\\;\sf{:\rightarrow\;\;area_{(in \: sq \: m)}\;=\;\bf{{0.11 {m}^{2} }\:}}\end{gathered}[\tex]</p><p></p><p>[tex]{\underline{\underline{\sf{\green{ Area \: of \: ABCD \: (margin) \: is \: 0.11m². }}}}}$