Ex 12.2

Ex 12.2, 1

Ex 12.2, 2

Ex 12.2, 3 Important

Ex 12.2, 4 Deleted for CBSE Board 2022 Exams You are here

Ex 12.2, 5 Important Deleted for CBSE Board 2022 Exams

Ex 12.2, 6 Deleted for CBSE Board 2022 Exams

Ex 12.2, 7 Important Deleted for CBSE Board 2022 Exams

Ex 12.2, 8 Important

Ex 12.2, 9

Ex 12.2, 10

Ex 12.2, 11 Important

Ex 12.2, 12

Ex 12.2, 13 Important Deleted for CBSE Board 2022 Exams

Ex 12.2, 14 (MCQ)

Last updated at May 29, 2018 by Teachoo

Ex 12.2, 4 A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : minor segment (Use = 3.14) Given that OA = OB = radius = 10 cm =90 Area of segment APB = Area of sector OAPB Area of AOB Area of sector OAPB = /(360 ) r2 = 90/360 3.14 (10)2 = 1/4 3.14 100 = 1/4 314 = 78.5 cm2 Area of AOB Now, AOB is a right triangle, where O = 90 having Base = OA & Height = OB Area of AOB = 1/2 Base Height = 1/2 OA OB = 1/2 10 10 = 50 cm2 Now, Area of segment APB = Area of sector OAPB Area of AOB = 78.5 50 = 28.5 cm2 Ex 12.2, 4(Method 1) A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (ii) major sector (Use = 3.14) Here, = 90 & radius = r = 10 cm Area of major sector = ((360 ))/(360 ) r2 = ((360 90))/360 3.14 (10)2 = 270/360 3.14 100 = 3/4 314 = 235.5 cm2 Ex 12.2, 4(Method 2) A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (ii) major sector (Use = 3.14) Area of major sector = Area of circle Area of sector OAPB Area of circle = r2 = 3.14 (10)2 = 3.14 100 = 314 cm2 Now, Area of major sector = Area of circle Area of sector OAPB = 314 78.5 = 235.5