Example 3 - Chapter 12 Class 10 Areas related to Circles (Term 1)

Transcript

Example 3 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° Area of segment AYB = Area of sector OAYB – Area of ΔOAB 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 Area Δ AOB = 1/2 × Base × Height We draw OM ⊥ AB ∴ ∠ OMB = ∠ OMA = 90° In Δ OMA & Δ OMB ∠ OMA = ∠ OMB OA = OB OM = OM ∴ Δ OMA ≅ Δ OMB ⇒ ∠ AOM = ∠ BOM ∴ ∠ AOM = ∠ BOM = 1/2 ∠ BOA Also, since Δ OMB ≅ Δ OMA ∴ BM = AM ⇒ BM = AM = 1/2 AB From (1) AM = 1/2AB 2AM = AB AB = 2AM Putting value of AM AB = 2 × √3/2 × 21 AB = √3 × 21 AB = 21√3 Now, Area of Δ AOB = 1/2 × Base × Height = 1/2 × AB × OM = 1/2 × 21√3 × 21/2 = (441√3)/4 Area of the segment AYB = Area of sector – area of ∆ 𝐴𝑂𝐵 = (462 – 441/4 √3 ) cm2