Answer
425.4k+ views
Hint: In the problem we need to find the shaded region so what we have to do is finding the area of right angled triangle in semicircle part by using the formula for area of triangle and then we have to find the area of semicircle. Now by subtracting the area of the right angled triangle from the area of the semicircle we will get the area of the shaded part.
Complete step by step answer:
We know that any angle made by the diameter QR in the semicircle is \[{{90}^{\circ }}\]
Hence \[\angle RPQ={{90}^{\circ }}\]
In the right angled \[\Delta RPQ\]
By Pythagoras we know that
\[R{{Q}^{2}}=P{{Q}^{2}}+P{{R}^{2}}\]
\[\begin{align}
& R{{Q}^{2}}={{24}^{2}}+{{7}^{2}} \\
& R{{Q}^{2}}=576+49 \\
& R{{Q}^{2}}=625 \\
& RQ=\sqrt{625} \\
& RQ=25cm \\
\end{align}\]
We know that RQ is the diameter of circle so the radius of circle is \[OQ=\dfrac{RQ}{2}=\dfrac{25}{2}cm\]
The area of right angled \[\Delta RPQ=\dfrac{1}{2}\times base\times height\]
The area of right angled \[\Delta RPQ=\dfrac{1}{2}\times RP\times PQ\]
The area of right angled \[\Delta RPQ=\dfrac{1}{2}\times 7\times 24\] \[=7\times 12=84c{{m}^{2}}\]. . . . . . . . (1)
We know that the area of semicircle is given by the formula \[=\dfrac{\pi {{r}^{2}}}{2}\]
\[=\dfrac{22}{7}\times \dfrac{25}{2}\times \dfrac{25}{2}\times \dfrac{1}{2}\]
\[=\dfrac{11\times 25\times 25}{28}\]
\[=\dfrac{6875}{28}c{{m}^{2}}\]. . . . . . . . . . . (2)
The area of shaded region is given by area of semicircle -area of right angled \[\Delta RPQ\]
\[=\dfrac{6875}{28}-84\]
\[=\dfrac{6875-2352}{28}\]
\[=\dfrac{4523}{28}=161.54c{{m}^{2}}\]
Hence the area of the shaded region is \[161.54c{{m}^{2}}\]
So, the correct answer is “Option B”.
Note: we know that the formula for area of triangle is given by \[\dfrac{1}{2}\times base\times height\] and formula for area of semicircle is given by \[\dfrac{\pi {{r}^{2}}}{2}\]. The diameter of the circle cuts the given circle into two equal parts in other words it divides into two semicircles. In this problem line RQ divides into two semicircles.
Complete step by step answer:
We know that any angle made by the diameter QR in the semicircle is \[{{90}^{\circ }}\]
Hence \[\angle RPQ={{90}^{\circ }}\]
In the right angled \[\Delta RPQ\]
By Pythagoras we know that
\[R{{Q}^{2}}=P{{Q}^{2}}+P{{R}^{2}}\]
\[\begin{align}
& R{{Q}^{2}}={{24}^{2}}+{{7}^{2}} \\
& R{{Q}^{2}}=576+49 \\
& R{{Q}^{2}}=625 \\
& RQ=\sqrt{625} \\
& RQ=25cm \\
\end{align}\]
We know that RQ is the diameter of circle so the radius of circle is \[OQ=\dfrac{RQ}{2}=\dfrac{25}{2}cm\]
The area of right angled \[\Delta RPQ=\dfrac{1}{2}\times base\times height\]
The area of right angled \[\Delta RPQ=\dfrac{1}{2}\times RP\times PQ\]
The area of right angled \[\Delta RPQ=\dfrac{1}{2}\times 7\times 24\] \[=7\times 12=84c{{m}^{2}}\]. . . . . . . . (1)
We know that the area of semicircle is given by the formula \[=\dfrac{\pi {{r}^{2}}}{2}\]
\[=\dfrac{22}{7}\times \dfrac{25}{2}\times \dfrac{25}{2}\times \dfrac{1}{2}\]
\[=\dfrac{11\times 25\times 25}{28}\]
\[=\dfrac{6875}{28}c{{m}^{2}}\]. . . . . . . . . . . (2)
The area of shaded region is given by area of semicircle -area of right angled \[\Delta RPQ\]
\[=\dfrac{6875}{28}-84\]
\[=\dfrac{6875-2352}{28}\]
\[=\dfrac{4523}{28}=161.54c{{m}^{2}}\]
Hence the area of the shaded region is \[161.54c{{m}^{2}}\]
So, the correct answer is “Option B”.
Note: we know that the formula for area of triangle is given by \[\dfrac{1}{2}\times base\times height\] and formula for area of semicircle is given by \[\dfrac{\pi {{r}^{2}}}{2}\]. The diameter of the circle cuts the given circle into two equal parts in other words it divides into two semicircles. In this problem line RQ divides into two semicircles.
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)