Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Important Area Questions
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Deleted for CBSE Board 2024 Exams
Question 8 Important Deleted for CBSE Board 2024 Exams
Question 9 Deleted for CBSE Board 2024 Exams
Question 10 Deleted for CBSE Board 2024 Exams
Question 11 Deleted for CBSE Board 2024 Exams
Question 12 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 Deleted for CBSE Board 2024 Exams
Question 15 Important Deleted for CBSE Board 2024 Exams You are here
Question 16 Important Deleted for CBSE Board 2024 Exams
Important Area Questions
Last updated at May 29, 2023 by Teachoo
Question 15 In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. Area of shaded region = Area of semicircle BEC (Area of quadrant ABDC Area of ABC) Area quadrant ABDC Radius = 14 cm Area of quadrant ABDC = 1/4 (area of circle) = 1/4 ( r2) = 1/4 22/7 (14)^2 = 1/4 22/7 14 14 = 154 cm2 Area triangle ABC Since ABDC is a quadrant BAC = 90 Hence ABC is a right triangle with Base AC & Height AB So, Area ABC = 1/2 Base Height = 1/2 AC AB = 1/2 14 14 = 7 14 = 98 cm2 Area semicircle BEC Here, Diameter = BC Finding BC first Since ABC is a right triangle Applying Pythagoras theorem in right triangle ABC (Hypotenuse)2 = (Height)2 + (Base)2 (BC)2 = (AB)2 + (AC)2 (BC)2 = (14)2 + (14)2 (BC)2 = (14)2 2 BC = ("(14)2 2" ) BC = ("(14)2" ) ("2" ) BC = 14 ("2" ) BC = 14 2 cm So, Diameter = BC = 14 2 cm Radius = r = /2 = /2 = (14 2)/2 = 7 2 cm Area of semicircle BEC = 1/2 Area of circle = 1/2 ( 2) = 1/2 22/7 (7 2)^2 = 1/2 22/7 7 7 2 2 = 1/2 22 7 2 = 154 cm2 Area of semi circle BEC = 154 cm2 Area of shaded region = Area of semicircle BEC (Area of quadrant ABDC Area of ABC) = 154 (154 98) = 154 154 + 98 = 98 cm2 Hence, area of shaded region = 98 cm2