NCERT Solutions, Chapter 12: Areas Related to Circles, Class 10 (Mathematics) Class 10 Notes | EduRev

Class 10 : NCERT Solutions, Chapter 12: Areas Related to Circles, Class 10 (Mathematics) Class 10 Notes | EduRev

 Page 1


 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
23 
  
  
  
Exercise 12.3  
Question 1:  
Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O 
is the centre of the circle. [?????? ?? =
22
7
] 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
Answer 1:  
It can be observed that RQ is the diameter of the circle. Therefore, ?RPQ will be 90º.  
By applying Pythagoras theorem in ?PQR,  
RP
2
 + PQ
2
 = RQ
2
  
(7)
2
 + (24)
2
 = RQ
2
  
  
Since RQ is the diameter of the circle, it divides the circle in two equal parts.  
  
Radius of circle,  
Page 2


 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
23 
  
  
  
Exercise 12.3  
Question 1:  
Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O 
is the centre of the circle. [?????? ?? =
22
7
] 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
Answer 1:  
It can be observed that RQ is the diameter of the circle. Therefore, ?RPQ will be 90º.  
By applying Pythagoras theorem in ?PQR,  
RP
2
 + PQ
2
 = RQ
2
  
(7)
2
 + (24)
2
 = RQ
2
  
  
Since RQ is the diameter of the circle, it divides the circle in two equal parts.  
  
Radius of circle,  
 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
24 
  
 
Area of shaded region = Area of semi-circle RPQOR - Area of ?PQR  
=  
=   
=  
 
 
 
 
 
 
 
 
 
 
 
 
 
Area of ?PQR    
  
  
cm 
2 
  
Page 3


 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
23 
  
  
  
Exercise 12.3  
Question 1:  
Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O 
is the centre of the circle. [?????? ?? =
22
7
] 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
Answer 1:  
It can be observed that RQ is the diameter of the circle. Therefore, ?RPQ will be 90º.  
By applying Pythagoras theorem in ?PQR,  
RP
2
 + PQ
2
 = RQ
2
  
(7)
2
 + (24)
2
 = RQ
2
  
  
Since RQ is the diameter of the circle, it divides the circle in two equal parts.  
  
Radius of circle,  
 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
24 
  
 
Area of shaded region = Area of semi-circle RPQOR - Area of ?PQR  
=  
=   
=  
 
 
 
 
 
 
 
 
 
 
 
 
 
Area of ?PQR    
  
  
cm 
2 
  
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
www.tiwariacademy.com  
25 
Question 2:  
Find the area of the shaded region in the given figure, if radii of the two concentric circles 
with centre O are 7 cm and 14 cm respectively and ?AOC = 40°. [?????? ?? =
22
7
] 
Answer 2:  
Page 4


 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
23 
  
  
  
Exercise 12.3  
Question 1:  
Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O 
is the centre of the circle. [?????? ?? =
22
7
] 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
Answer 1:  
It can be observed that RQ is the diameter of the circle. Therefore, ?RPQ will be 90º.  
By applying Pythagoras theorem in ?PQR,  
RP
2
 + PQ
2
 = RQ
2
  
(7)
2
 + (24)
2
 = RQ
2
  
  
Since RQ is the diameter of the circle, it divides the circle in two equal parts.  
  
Radius of circle,  
 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
24 
  
 
Area of shaded region = Area of semi-circle RPQOR - Area of ?PQR  
=  
=   
=  
 
 
 
 
 
 
 
 
 
 
 
 
 
Area of ?PQR    
  
  
cm 
2 
  
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
www.tiwariacademy.com  
25 
Question 2:  
Find the area of the shaded region in the given figure, if radii of the two concentric circles 
with centre O are 7 cm and 14 cm respectively and ?AOC = 40°. [?????? ?? =
22
7
] 
Answer 2:  
 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
26 
Question 3:  
Find the area of the shaded region in the given figure, if ABCD is a square of side 14  
cm and APD and BPC are semicircles. [?????? ?? =
22
7
]  
 
 
Answer 3:  
It can be observed from the figure that the radius of each semi-circle is 7 cm.  
 
 
 
 
 
  
 
 
 
 
 
 
  
Area of square ABCD = (Side)
2
 = (14)
2
 = 196 cm
2
  
Area of the shaded region   
= Area of square ABCD - Area of semi-circle APD - Area of semi-circle BPC  
= 196 - 77 - 77 = 196 - 154 = 42 cm
2
  
Area of each semi-circle =  
Page 5


 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
23 
  
  
  
Exercise 12.3  
Question 1:  
Find the area of the shaded region in the given figure, if PQ = 24 cm, PR = 7 cm and O 
is the centre of the circle. [?????? ?? =
22
7
] 
 
  
 
 
 
 
 
 
 
 
 
 
 
 
Answer 1:  
It can be observed that RQ is the diameter of the circle. Therefore, ?RPQ will be 90º.  
By applying Pythagoras theorem in ?PQR,  
RP
2
 + PQ
2
 = RQ
2
  
(7)
2
 + (24)
2
 = RQ
2
  
  
Since RQ is the diameter of the circle, it divides the circle in two equal parts.  
  
Radius of circle,  
 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
24 
  
 
Area of shaded region = Area of semi-circle RPQOR - Area of ?PQR  
=  
=   
=  
 
 
 
 
 
 
 
 
 
 
 
 
 
Area of ?PQR    
  
  
cm 
2 
  
Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
www.tiwariacademy.com  
25 
Question 2:  
Find the area of the shaded region in the given figure, if radii of the two concentric circles 
with centre O are 7 cm and 14 cm respectively and ?AOC = 40°. [?????? ?? =
22
7
] 
Answer 2:  
 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
26 
Question 3:  
Find the area of the shaded region in the given figure, if ABCD is a square of side 14  
cm and APD and BPC are semicircles. [?????? ?? =
22
7
]  
 
 
Answer 3:  
It can be observed from the figure that the radius of each semi-circle is 7 cm.  
 
 
 
 
 
  
 
 
 
 
 
 
  
Area of square ABCD = (Side)
2
 = (14)
2
 = 196 cm
2
  
Area of the shaded region   
= Area of square ABCD - Area of semi-circle APD - Area of semi-circle BPC  
= 196 - 77 - 77 = 196 - 154 = 42 cm
2
  
Area of each semi-circle =  
 Mathematics 
(www.tiwariacademy.com) 
(Chapter – 12) (Areas Related to Circles)  
(Class – X) 
 
www.tiwariacademy.com  
27 
Question 4:  
Find the area of the shaded region in the given figure, where a circular arc of radius 6 
cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre. 
[?????? ?? =
22
7
]  
  
 
 
Answer 4:  
We know that each interior angle of an equilateral triangle is of measure 60°.  
 
 
 
 
 
 
 
 
 
 
 
 
  
  
Area of sector OCDE  
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