Answer
Verified
477.9k+ views
Hint – In this question we have to find the area of the shaded region, so using the concept that if the smaller circle area is removed from the larger circle’s area then eventually the only left area is the required shaded region.
Complete step-by-step answer:
As we know that the area (A) of the circle is $\pi {r^2}$ (where r is the radius of the small circle).
Now as we know that diameter (d) of a circle is twice the radius.
$
\Rightarrow d = 2r \\
\Rightarrow r = \dfrac{d}{2} \\
$
So substitute this value in the formula of area of small circle we have
$ \Rightarrow A = \pi {\left( {\dfrac{d}{2}} \right)^2} = \pi \dfrac{{{d^2}}}{4}$……………… (1)
Now it is given that AB and CD are the two diameters of a circle perpendicular to each other.
$ \Rightarrow AB = CD$ (Diameter of the circle).
And we know half of the diameter is the radius of the circle.
$ \Rightarrow OA = OB = OC = OD = \dfrac{{AB}}{2} = \dfrac{{CD}}{2}$ (Radius of the circle).
Now it is given that OD = 7 unit and it is the diameter (${d_1}$) of the smaller circle.
So the area (${A_1}$) of the smaller circle from equation (1)
$ \Rightarrow {A_1} = \pi {\left( {\dfrac{{{d_1}}}{2}} \right)^2} = \pi \dfrac{{{d_1}^2}}{4} = \pi \dfrac{{{7^2}}}{4} = \dfrac{{49\pi }}{4}$.
And the diameter (${d_2}$) of the big circle is $\left( {2 \times 7} \right) = 14$ unit.
So the area (${A_2}$) of the big circle is
$ \Rightarrow {A_2} = \pi {\left( {\dfrac{{{d_2}}}{2}} \right)^2} = \pi \dfrac{{{d_2}^2}}{4} = \pi \dfrac{{{{\left( {14} \right)}^2}}}{4} = \dfrac{{196\pi }}{4}$
So the area (${A_3}$) of the shaded portion is
= Area of big circle – area of smaller circle.
$ \Rightarrow {A_3} = \dfrac{{196\pi }}{4} - \dfrac{{49\pi }}{4} = \dfrac{{147\pi }}{4}$
Now substitute $\left[ {\pi = \dfrac{{22}}{7}} \right]$ we have,
$ \Rightarrow {A_3} = \dfrac{{147\pi }}{4} = \dfrac{{147}}{4}\left( {\dfrac{{22}}{7}} \right) = \dfrac{{21 \times 11}}{2} = \dfrac{{231}}{2}$ Sq. units.
So, the area of the shaded portion is $\dfrac{{231}}{2}$ sq. units.
So, this is the required answer.
Note – Whenever we face such types of problems the key concept is simply to have the gist of the basic formula of area of circle and the basics of obtaining radius of circle using diameter. The diagrammatic representation of the figure always helps in better understanding of the geometry involved for the shaded area.
Complete step-by-step answer:
As we know that the area (A) of the circle is $\pi {r^2}$ (where r is the radius of the small circle).
Now as we know that diameter (d) of a circle is twice the radius.
$
\Rightarrow d = 2r \\
\Rightarrow r = \dfrac{d}{2} \\
$
So substitute this value in the formula of area of small circle we have
$ \Rightarrow A = \pi {\left( {\dfrac{d}{2}} \right)^2} = \pi \dfrac{{{d^2}}}{4}$……………… (1)
Now it is given that AB and CD are the two diameters of a circle perpendicular to each other.
$ \Rightarrow AB = CD$ (Diameter of the circle).
And we know half of the diameter is the radius of the circle.
$ \Rightarrow OA = OB = OC = OD = \dfrac{{AB}}{2} = \dfrac{{CD}}{2}$ (Radius of the circle).
Now it is given that OD = 7 unit and it is the diameter (${d_1}$) of the smaller circle.
So the area (${A_1}$) of the smaller circle from equation (1)
$ \Rightarrow {A_1} = \pi {\left( {\dfrac{{{d_1}}}{2}} \right)^2} = \pi \dfrac{{{d_1}^2}}{4} = \pi \dfrac{{{7^2}}}{4} = \dfrac{{49\pi }}{4}$.
And the diameter (${d_2}$) of the big circle is $\left( {2 \times 7} \right) = 14$ unit.
So the area (${A_2}$) of the big circle is
$ \Rightarrow {A_2} = \pi {\left( {\dfrac{{{d_2}}}{2}} \right)^2} = \pi \dfrac{{{d_2}^2}}{4} = \pi \dfrac{{{{\left( {14} \right)}^2}}}{4} = \dfrac{{196\pi }}{4}$
So the area (${A_3}$) of the shaded portion is
= Area of big circle – area of smaller circle.
$ \Rightarrow {A_3} = \dfrac{{196\pi }}{4} - \dfrac{{49\pi }}{4} = \dfrac{{147\pi }}{4}$
Now substitute $\left[ {\pi = \dfrac{{22}}{7}} \right]$ we have,
$ \Rightarrow {A_3} = \dfrac{{147\pi }}{4} = \dfrac{{147}}{4}\left( {\dfrac{{22}}{7}} \right) = \dfrac{{21 \times 11}}{2} = \dfrac{{231}}{2}$ Sq. units.
So, the area of the shaded portion is $\dfrac{{231}}{2}$ sq. units.
So, this is the required answer.
Note – Whenever we face such types of problems the key concept is simply to have the gist of the basic formula of area of circle and the basics of obtaining radius of circle using diameter. The diagrammatic representation of the figure always helps in better understanding of the geometry involved for the shaded area.
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