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 OCDERead More

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