Example 2 - Chapter 11 Class 10 Areas related to Circles

Transcript

Example 2 Find the area of the segment AYB shown in figure, if radius of the circle is 21 cm and ∠AOB = 120°. (Use π = 22/7 ). In a given circle, Radius (r) = 21 cm And, 𝜽 = 120° Now, Area of segment AYB = Area of sector OAYB – Area of ΔOAB Finding Area of sector OAYB Area of sector OAYB = 𝜃/360× 𝜋𝑟2 = 120/360 × 22/7×(21)2 = 1/3×22/7 × 21 × 21 = 22 × 21 = 462 cm2 Finding area of Δ AOB We draw OM ⊥ AB ∴ ∠ OMB = ∠ OMA = 90° And, by symmetry M is the mid-point of AB ∴ BM = AM = 1/2 AB In right triangle Δ OMA sin O = (side opposite to angle O)/Hypotenuse sin 𝟔𝟎° = 𝐀𝑴/𝑨𝑶 √3/2=𝐴𝑀/21 √3/2 × 21 = AM AM = √𝟑/𝟐 × 21 In right triangle Δ OMA cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 𝟔𝟎° = 𝑶𝑴/𝑨𝑶 1/2=𝑂𝑀/21 21/2 = OM OM = 𝟐𝟏/𝟐 From (1) AM = 𝟏/𝟐AB 2AM = AB AB = 2AM Putting value of AM AB = 2 × √3/2 × 21 AB = √3 × 21 AB = 21√𝟑 cm Now, Area of Δ AOB = 1/2 × Base × Height = 𝟏/𝟐 × AB × OM = 1/2 × 21√3 × 21/2 = (𝟒𝟒𝟏√𝟑)/𝟒 cm2 Therefore, Area of the segment AYB = Area of sector – Area of ∆ 𝐴𝑂𝐵 = (462 – 𝟒𝟒𝟏/𝟒 √𝟑 ) cm2