








Get live Maths 1-on-1 Classs - Class 6 to 12
Miscellaneous
Misc 1 (ii) Important
Misc 2 Deleted for CBSE Board 2023 Exams
Misc 3
Misc 4 Important
Misc 5 Important
Misc 6 Important
Misc 7
Misc 8
Misc 9 Important
Misc 10
Misc 11 Important
Misc 12 Important Deleted for CBSE Board 2023 Exams
Misc 13
Misc 14 Important
Misc 15 Deleted for CBSE Board 2023 Exams You are here
Misc 16 (MCQ)
Misc 17 (MCQ) Important
Misc 18 (MCQ) Deleted for CBSE Board 2023 Exams
Misc 19 (MCQ) Important Deleted for CBSE Board 2023 Exams
Miscellaneous
Last updated at March 30, 2023 by Teachoo
Misc 15 Find the area of the region { , : 2 4 , 4 2+ 4 2 9} Drawing figures Now, our figure will be Circle is 4 2+ 4 2 9 2+ 2 9 4 2+ 2 3 2 2 Radius of circle = 3 2 So, Point C ( 3 2 , 0) Finding point of intersection A & B Solving 2 =4 (1) & 4 2+ 4 2= 9 (2) Putting (1) in (2) 4 2+ 4 2= 9 4 2+ 4 4 =9 4 2+16 9=0 4 2+18 2 9=0 2 (2 +9) 1(2 +9)=0 2 1 2 +9 =0 So, = 1 2 & = 9 2 Area OAC Area OAC = Area ACD + Area OAD Area ACD Area ACD = 1 2 3 2 Equation of circle 2 + 2 = 9 4 2 = 9 4 2 = 9 4 2 Therefore Area ACD = 1 2 3 2 9 4 2 = 1 2 3 2 3 2 2 2 = 2 3 2 2 2 + 3 2 2 2 sin 1 3 2 1 2 3 2 = 2 9 4 2 + 9 8 sin 1 2 3 1 2 3 2 = 3 2 2 9 4 3 2 2 + 9 8 sin 1 2 3 2 3 1 2 2 9 4 1 2 2 + 9 8 sin 1 2 1 2 3 = 3 4 9 4 9 4 + 9 8 sin 1 1 1 4 9 4 1 4 + 9 8 sin 1 1 3 = 9 8 sin 1 1 2 4 9 8 sin 1 1 3 = 9 16 2 4 9 8 sin 1 1 3 Area OAD Area OAD = 0 1 2 Equation of parabola 2 =4 = 4 Therefore Area OAD = 0 1 2 4 =2 0 1 2 =2 0 1 2 1 2 =2 3 2 3 2 0 1 2 = 4 3 [ 3 2 ] 0 1 2 = 4 3 1 2 3 2 0 3 2 = 4 3 1 2 2 = 4 3 1 2 2 2 2 = 2 3 Area OAC = Area ACD + Area OAD = 9 16 2 4 9 8 sin 1 1 3 + 2 3 = 9 16 9 8 sin 1 1 3 + 2 12 Required Area = 2 Area OAC = 2 9 16 9 8 sin 1 1 3 + 2 12 = 9 8 9 4 sin 1 1 3 + 2 6 = 9 8 9 4 sin 1 1 3 + 2 6 2 2 = 9 8 9 4 sin 1 1 3 + 2 6 2 = +