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# 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
(Chapter – 12) (Areas Related to Circles)
(Class – X)

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
]

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.

Page 2

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)

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
]

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.

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)

24

Area of shaded region = Area of semi-circle RPQOR - Area of ?PQR
=
=
=

Area of ?PQR

cm
2

Page 3

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)

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
]

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.

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)

24

Area of shaded region = Area of semi-circle RPQOR - Area of ?PQR
=
=
=

Area of ?PQR

cm
2

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)
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
]
Page 4

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)

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
]

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.

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)

24

Area of shaded region = Area of semi-circle RPQOR - Area of ?PQR
=
=
=

Area of ?PQR

cm
2

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)
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
]
Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)

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
]

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 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
(Chapter – 12) (Areas Related to Circles)
(Class – X)

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
]

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.

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)

24

Area of shaded region = Area of semi-circle RPQOR - Area of ?PQR
=
=
=

Area of ?PQR

cm
2

Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)
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
]
Mathematics
(Chapter – 12) (Areas Related to Circles)
(Class – X)

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
]

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 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
(Chapter – 12) (Areas Related to Circles)
(Class – X)

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
]

We know that each interior angle of an equilateral triangle is of measure 60°.

Area of sector OCDE
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