Answer
Verified
478.8k+ views
Hint: To find the area of the shaded region in the given figure we will subtract the area of the circle having radius 1 cm and the two semi-circle having radius 2 cm from the bigger semi-circle of radius 4 cm. Then we will get the area of the shaded region.
Complete step-by-step answer:
We have been given the figure as follows:
From the given figure, we can say that the area of the shaded region is equal to the area obtained by subtracting the area of the circle with C as center and the two semi-circles with AM and MB as diameter from the bigger semi-circle having AB as diameter.
We know that area of a circle having D as diameter is as follows:
Area \[=\dfrac{\pi {{D}^{2}}}{4}\]
So, area of semi-circle \[=\dfrac{1}{2}\times \] area of circle
\[\begin{align}
& =\dfrac{1}{2}\times \dfrac{\pi {{D}^{2}}}{4} \\
& =\dfrac{\pi {{D}^{2}}}{8} \\
\end{align}\]
Now the area of semi-circle having AM = 4 cm as diameter \[=\dfrac{\pi {{D}^{2}}}{8}\]
\[=\dfrac{\pi \times {{\left( 4 \right)}^{2}}}{8}=\dfrac{\pi \times 16}{8}=2\pi c{{m}^{2}}\]
Area of semi-circle having MB = 4 cm as diameter \[=\dfrac{\pi {{D}^{2}}}{8}\]
\[=\dfrac{\pi \times 4}{8}=\dfrac{\pi \times 16}{8}=2\pi c{{m}^{2}}\]
Area of circle having C as center and diameter \[=\left( 2\times 1 \right)cm\] is \[=\dfrac{\pi {{D}^{2}}}{4}\]
\[=\dfrac{\pi \times {{\left( 2 \right)}^{2}}}{4}=\pi c{{m}^{2}}\]
Area of semicircle having AB = 8 cm as diameter \[=\dfrac{\pi {{D}^{2}}}{8}\]
\[=\dfrac{\pi \times {{8}^{2}}}{8}=\pi \times 8=8\pi c{{m}^{2}}\]
So the area of the shaded region = area of semi-circle having AB as diameter – (area of the circle having C as center + 2 area of semi-circle with AM as diameter)
\[=8\pi -\left( \pi +2\times 2\pi \right)=8\pi -5\pi =3\pi c{{m}^{2}}\]
Also, \[\pi =\dfrac{22}{7}\] so by substituting the value of \[\pi \] we get as follows:
Area of shaded region \[=3\pi =3\times \dfrac{22}{7}=9.43c{{m}^{2}}(approx)\]
Therefore, the area of the shaded region is \[9.43c{{m}^{2}}(approx)\].
Note: Be careful while calculating the area as we have been given the diameter of some circle and radius of a circle so accordingly use the formula to calculate the area.
Complete step-by-step answer:
We have been given the figure as follows:
From the given figure, we can say that the area of the shaded region is equal to the area obtained by subtracting the area of the circle with C as center and the two semi-circles with AM and MB as diameter from the bigger semi-circle having AB as diameter.
We know that area of a circle having D as diameter is as follows:
Area \[=\dfrac{\pi {{D}^{2}}}{4}\]
So, area of semi-circle \[=\dfrac{1}{2}\times \] area of circle
\[\begin{align}
& =\dfrac{1}{2}\times \dfrac{\pi {{D}^{2}}}{4} \\
& =\dfrac{\pi {{D}^{2}}}{8} \\
\end{align}\]
Now the area of semi-circle having AM = 4 cm as diameter \[=\dfrac{\pi {{D}^{2}}}{8}\]
\[=\dfrac{\pi \times {{\left( 4 \right)}^{2}}}{8}=\dfrac{\pi \times 16}{8}=2\pi c{{m}^{2}}\]
Area of semi-circle having MB = 4 cm as diameter \[=\dfrac{\pi {{D}^{2}}}{8}\]
\[=\dfrac{\pi \times 4}{8}=\dfrac{\pi \times 16}{8}=2\pi c{{m}^{2}}\]
Area of circle having C as center and diameter \[=\left( 2\times 1 \right)cm\] is \[=\dfrac{\pi {{D}^{2}}}{4}\]
\[=\dfrac{\pi \times {{\left( 2 \right)}^{2}}}{4}=\pi c{{m}^{2}}\]
Area of semicircle having AB = 8 cm as diameter \[=\dfrac{\pi {{D}^{2}}}{8}\]
\[=\dfrac{\pi \times {{8}^{2}}}{8}=\pi \times 8=8\pi c{{m}^{2}}\]
So the area of the shaded region = area of semi-circle having AB as diameter – (area of the circle having C as center + 2 area of semi-circle with AM as diameter)
\[=8\pi -\left( \pi +2\times 2\pi \right)=8\pi -5\pi =3\pi c{{m}^{2}}\]
Also, \[\pi =\dfrac{22}{7}\] so by substituting the value of \[\pi \] we get as follows:
Area of shaded region \[=3\pi =3\times \dfrac{22}{7}=9.43c{{m}^{2}}(approx)\]
Therefore, the area of the shaded region is \[9.43c{{m}^{2}}(approx)\].
Note: Be careful while calculating the area as we have been given the diameter of some circle and radius of a circle so accordingly use the formula to calculate the area.
Recently Updated Pages
10 Examples of Evaporation in Daily Life with Explanations
10 Examples of Diffusion in Everyday Life
1 g of dry green algae absorb 47 times 10 3 moles of class 11 chemistry CBSE
If x be real then the maximum value of 5 + 4x 4x2 will class 10 maths JEE_Main
If the coordinates of the points A B and C be 443 23 class 10 maths JEE_Main
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Who was the leader of the Bolshevik Party A Leon Trotsky class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Which is the largest saltwater lake in India A Chilika class 8 social science CBSE
Ghatikas during the period of Satavahanas were aHospitals class 6 social science CBSE