# Ex 8.1, 7

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Ex 8.1, 7 Find the area of the smaller part of the circle 𝑥2+𝑦2= 𝑎2 cut off by the line 𝑥= 𝑎 2. Equation of Given Circle is 𝑥2+ 𝑦2= 𝑎2 ∴ Radius , 𝑟 = 𝑎 So, a is positive ∴ 𝑥= 𝑎 2 will lie on positive side of x-axis Let PQ represent the line 𝑥= 𝑎 2 We have to find Area of APQ Area APQ = 2 × Area APR = 2 × We know that, 𝑥2+ 𝑦2= 𝑎2 𝑦2= 𝑎2− 𝑥2 𝑦=± 𝑎2− 𝑥2 Since , APR is in 1st Quadrant 𝑦= 𝑎2− 𝑥2 ∴ Area of APQ = 2 × = 2 × = 2 12𝑥 𝑎2− 𝑥2+ 𝑎2𝑎 sin−1 𝑥𝑎 𝑎 2𝑎 = 2 𝑎2 𝑎2− 𝑎2+ 𝑎22 sin−1 𝑎𝑎− 𝑎2 2 𝑎2− 𝑎 22− 𝑎22 sin−1 𝑎 2𝑎 = 2 0+ 𝑎22 sin−1 1− 𝑎2 2 𝑎2− 𝑎22− 𝑎22 sin−1 1 2 = 2 𝑎22𝑠𝑖 𝑛−1 1− 𝑎22 sin−1 1 2− 𝑎2 2 2𝑎2 − 𝑎22 = 2 𝑎22 sin−1 1− sin−1 1 2− 𝑎2 2 𝑎2 2 = 2 𝑎22 𝜋2− 𝜋4− 𝑎2 2 × 𝑎 2 = 2 𝑎22 𝜋2− 𝜋4− 𝑎22.2 = 2 𝑎22 𝜋4− 𝑎24 = 2 𝑎24 𝜋2−1 = 𝑎22 𝜋2−1 ∴ Required Area = 𝒂𝟐𝟐 𝝅𝟐−𝟏 Square units

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