Answer
453.6k+ views
Hint:- The side length of the square is equal to the diameter of the semicircle.
Given , the side length of the square is 14cm. Now, the shaded region as well as the semicircles are inside the square. Hence, the total area will comprise the area of the shaded region and the semicircles.
i.e.
${\text{Area of square = area of semicircle APC + area of shaded region + area of semicircle BPD}}$ -(1)
The area of the square is equal to \[{({\text{side length)}}^2}\] .
Area of Square ABCD $ = {\text{ }}{\left( {14{\text{cm}}} \right)^2}$
${\text{ = 196 c}}{{\text{m}}^2}$ -(2)
Now, the semicircles APC and BPD.
Side AC = diameter of the semicircle APC = diameter of the semicircle BPD
Area of semicircle = $\dfrac{\pi }{8} \times {\left( {{\text{diameter}}} \right)^2}$
Area of semicircle APC =$\dfrac{\pi }{8} \times {\left( {14{\text{cm}}} \right)^2}$
Taking $\pi $ = 3.141, we get
Area of semicircle APC = $76.97{\text{ c}}{{\text{m}}^2}$ -(3)
As the diameter of semicircle APC and BPD are equal, the area will also be equal.
So, area of semicircle BPD = $76.97{\text{ c}}{{\text{m}}^2}$ -(4)
Putting the values of areas of square and semi circles from equations (2),(3) and (4) into equation (1).We get,
$
\Rightarrow 196{\text{ = 76}}{\text{.97 + area of shaded region + 76}}{\text{.97}} \\
\Rightarrow {\text{area of shaded region = 196 - 76}}{\text{.97 - 76}}{\text{.97}} \\
{\text{ = 42}}{\text{.06 c}}{{\text{m}}^2} \\
$
So, the answer is 42.06 square cm.
Note:- In these types of questions, for finding the solution a basic relation is considered i.e. outer area is equated to sum of the area’s of inner portions. Always remember to make sure that the area of each portion is present in the same square unit.
Given , the side length of the square is 14cm. Now, the shaded region as well as the semicircles are inside the square. Hence, the total area will comprise the area of the shaded region and the semicircles.
i.e.
${\text{Area of square = area of semicircle APC + area of shaded region + area of semicircle BPD}}$ -(1)
The area of the square is equal to \[{({\text{side length)}}^2}\] .
![seo images](https://www.vedantu.com/question-sets/27ef2912-a69f-4e27-83ba-5fadd2da12172949996024577368329.png)
Area of Square ABCD $ = {\text{ }}{\left( {14{\text{cm}}} \right)^2}$
${\text{ = 196 c}}{{\text{m}}^2}$ -(2)
Now, the semicircles APC and BPD.
![seo images](https://www.vedantu.com/question-sets/06acd833-150a-4fde-859a-b36f0f2bfbd25612472818532995222.png)
Side AC = diameter of the semicircle APC = diameter of the semicircle BPD
Area of semicircle = $\dfrac{\pi }{8} \times {\left( {{\text{diameter}}} \right)^2}$
Area of semicircle APC =$\dfrac{\pi }{8} \times {\left( {14{\text{cm}}} \right)^2}$
Taking $\pi $ = 3.141, we get
Area of semicircle APC = $76.97{\text{ c}}{{\text{m}}^2}$ -(3)
As the diameter of semicircle APC and BPD are equal, the area will also be equal.
So, area of semicircle BPD = $76.97{\text{ c}}{{\text{m}}^2}$ -(4)
Putting the values of areas of square and semi circles from equations (2),(3) and (4) into equation (1).We get,
$
\Rightarrow 196{\text{ = 76}}{\text{.97 + area of shaded region + 76}}{\text{.97}} \\
\Rightarrow {\text{area of shaded region = 196 - 76}}{\text{.97 - 76}}{\text{.97}} \\
{\text{ = 42}}{\text{.06 c}}{{\text{m}}^2} \\
$
So, the answer is 42.06 square cm.
Note:- In these types of questions, for finding the solution a basic relation is considered i.e. outer area is equated to sum of the area’s of inner portions. Always remember to make sure that the area of each portion is present in the same square unit.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)