Answer
Verified
377.7k+ views
Hint: We have to find the area of the circular sheet, area of the two small circles and the area of the rectangle. We will be using the formulas $A=\pi {{r}^{2}}$ and $A=l\times b$ to find the area of the circle and rectangle respectively. To find the area of the remaining sheet, we have to subtract the sum of areas of the two small circles and the rectangle from the area of the circular sheet.
Complete step by step answer:
We are given the radius of the circular sheet is 14 cm.
Let us find the area of the circular sheet. We know that area of a circle is given by
$A=\pi {{r}^{2}}$
where r is the radius of the circle.
Let us assume the area of the circular sheet as ${{A}_{1}}$ . We have to substitute the values in the above formula.
$\begin{align}
& \Rightarrow {{A}_{1}}=\pi \times {{14}^{2}} \\
& \Rightarrow {{A}_{1}}=14\times 14\times \dfrac{22}{7} \\
\end{align}$
Let us cancel the common factors.
$\begin{align}
& \Rightarrow {{A}_{1}}={{\require{cancel}\cancel{14}}^{2}}\times 14\times \dfrac{22}{\require{cancel}\cancel{7}} \\
& \Rightarrow {{A}_{1}}=2\times 14\times 22 \\
& \Rightarrow {{A}_{1}}=616\text{ }c{{m}^{2}} \\
\end{align}$
We are given that two circles of radius 3.5cm are removed. Let us find the area of one small circle.
$\Rightarrow {{A}_{2}}=\pi {{\left( 3.5 \right)}^{2}}$
We can write 3.5 as $\dfrac{35}{10}$ .
$\begin{align}
& \Rightarrow {{A}_{2}}=\dfrac{22}{7}\times {{\left( \dfrac{35}{10} \right)}^{2}} \\
& \Rightarrow {{A}_{2}}=\dfrac{22}{7}\times \dfrac{35}{10}\times \dfrac{35}{10} \\
\end{align}$
Let us cancel the common factors.
$\begin{align}
& \Rightarrow {{A}_{2}}=\dfrac{{{\require{cancel}\cancel{22}}^{11}}}{\require{cancel}\cancel{7}}\times \dfrac{{{\require{cancel}\cancel{35}}^{\require{cancel}\cancel{5}}}}{{{\require{cancel}\cancel{10}}^{\require{cancel}\cancel{2}}}}\times \dfrac{{{\require{cancel}\cancel{35}}^{7}}}{{{\require{cancel}\cancel{10}}^{2}}} \\
& \Rightarrow {{A}_{2}}=11\times \dfrac{7}{2} \\
& \Rightarrow {{A}_{2}}=\dfrac{77}{2} \\
\end{align}$
$\Rightarrow {{A}_{2}}=38.5\text{ }c{{m}^{2}}$
We are also given that a rectangle of length 3 cm and breadth 1 cm is also removed. Let us find the area of the rectangle. We know that area of a rectangle is given by
$A=l\times b$
where l is the length and b is the breadth of the rectangle.
Therefore, area of the given rectangle is
$\begin{align}
& \Rightarrow {{A}_{3}}=3\times 1 \\
& \Rightarrow {{A}_{3}}=3\text{ }c{{m}^{2}} \\
\end{align}$
We have to find the area of the remaining sheet. We can obtain this area by subtracting the sum of areas of the two small circles and the rectangle from the area of the circular sheet.
$\text{Area of the remaining sheet}=\text{ Area of circular sheet}-\left( \text{Area of two small circle}+\text{Area of rectangle} \right)$We can write the areas as
\[\Rightarrow \text{Area of the remaining sheet}={{A}_{1}}-\left( {{A}_{2}}+{{A}_{2}}+{{A}_{3}} \right)\]
Let us substitute the values.
\[\begin{align}
& \Rightarrow \text{Area of the remaining sheet}=616\text{ }-\left( 38.5+38.5+3 \right) \\
& \Rightarrow \text{Area of the remaining sheet}=616\text{ }-\left( 77+3 \right) \\
& \Rightarrow \text{Area of the remaining sheet}=616\text{ }-80 \\
& \Rightarrow \text{Area of the remaining sheet}=536\text{ c}{{\text{m}}^{2}} \\
\end{align}\]
Therefore, the area of the remaining sheet is 536 $\text{c}{{\text{m}}^{2}}$ .
Note: Students must be thorough with the areas of shapes like circles, squares, rectangles and triangles. They have a chance of making a mistake by writing the area of the circle as $2\pi r$ or the area of the rectangle as $2\left( l+b \right)$ . These are the perimeters of circle and rectangle respectively. Students have a chance of making a mistake by finding the remaining area of the sheet by subtracting the difference of areas of the two small circles and the rectangle from the area of the circular sheet.
Complete step by step answer:
We are given the radius of the circular sheet is 14 cm.
Let us find the area of the circular sheet. We know that area of a circle is given by
$A=\pi {{r}^{2}}$
where r is the radius of the circle.
Let us assume the area of the circular sheet as ${{A}_{1}}$ . We have to substitute the values in the above formula.
$\begin{align}
& \Rightarrow {{A}_{1}}=\pi \times {{14}^{2}} \\
& \Rightarrow {{A}_{1}}=14\times 14\times \dfrac{22}{7} \\
\end{align}$
Let us cancel the common factors.
$\begin{align}
& \Rightarrow {{A}_{1}}={{\require{cancel}\cancel{14}}^{2}}\times 14\times \dfrac{22}{\require{cancel}\cancel{7}} \\
& \Rightarrow {{A}_{1}}=2\times 14\times 22 \\
& \Rightarrow {{A}_{1}}=616\text{ }c{{m}^{2}} \\
\end{align}$
We are given that two circles of radius 3.5cm are removed. Let us find the area of one small circle.
$\Rightarrow {{A}_{2}}=\pi {{\left( 3.5 \right)}^{2}}$
We can write 3.5 as $\dfrac{35}{10}$ .
$\begin{align}
& \Rightarrow {{A}_{2}}=\dfrac{22}{7}\times {{\left( \dfrac{35}{10} \right)}^{2}} \\
& \Rightarrow {{A}_{2}}=\dfrac{22}{7}\times \dfrac{35}{10}\times \dfrac{35}{10} \\
\end{align}$
Let us cancel the common factors.
$\begin{align}
& \Rightarrow {{A}_{2}}=\dfrac{{{\require{cancel}\cancel{22}}^{11}}}{\require{cancel}\cancel{7}}\times \dfrac{{{\require{cancel}\cancel{35}}^{\require{cancel}\cancel{5}}}}{{{\require{cancel}\cancel{10}}^{\require{cancel}\cancel{2}}}}\times \dfrac{{{\require{cancel}\cancel{35}}^{7}}}{{{\require{cancel}\cancel{10}}^{2}}} \\
& \Rightarrow {{A}_{2}}=11\times \dfrac{7}{2} \\
& \Rightarrow {{A}_{2}}=\dfrac{77}{2} \\
\end{align}$
$\Rightarrow {{A}_{2}}=38.5\text{ }c{{m}^{2}}$
We are also given that a rectangle of length 3 cm and breadth 1 cm is also removed. Let us find the area of the rectangle. We know that area of a rectangle is given by
$A=l\times b$
where l is the length and b is the breadth of the rectangle.
Therefore, area of the given rectangle is
$\begin{align}
& \Rightarrow {{A}_{3}}=3\times 1 \\
& \Rightarrow {{A}_{3}}=3\text{ }c{{m}^{2}} \\
\end{align}$
We have to find the area of the remaining sheet. We can obtain this area by subtracting the sum of areas of the two small circles and the rectangle from the area of the circular sheet.
$\text{Area of the remaining sheet}=\text{ Area of circular sheet}-\left( \text{Area of two small circle}+\text{Area of rectangle} \right)$We can write the areas as
\[\Rightarrow \text{Area of the remaining sheet}={{A}_{1}}-\left( {{A}_{2}}+{{A}_{2}}+{{A}_{3}} \right)\]
Let us substitute the values.
\[\begin{align}
& \Rightarrow \text{Area of the remaining sheet}=616\text{ }-\left( 38.5+38.5+3 \right) \\
& \Rightarrow \text{Area of the remaining sheet}=616\text{ }-\left( 77+3 \right) \\
& \Rightarrow \text{Area of the remaining sheet}=616\text{ }-80 \\
& \Rightarrow \text{Area of the remaining sheet}=536\text{ c}{{\text{m}}^{2}} \\
\end{align}\]
Therefore, the area of the remaining sheet is 536 $\text{c}{{\text{m}}^{2}}$ .
Note: Students must be thorough with the areas of shapes like circles, squares, rectangles and triangles. They have a chance of making a mistake by writing the area of the circle as $2\pi r$ or the area of the rectangle as $2\left( l+b \right)$ . These are the perimeters of circle and rectangle respectively. Students have a chance of making a mistake by finding the remaining area of the sheet by subtracting the difference of areas of the two small circles and the rectangle from the area of the circular sheet.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE