Ex 6.3, 7

Transcript

Ex 6.4, 7 A chord of a circle of radius 𝑟 subtends an angle of 60^∘ at the centre of the circle. Show that the area of the corresponding minor segment of the circle is equal to 𝜋𝑟^2 (1/6−√3/4). Let the minor segment be APB Given that 𝜃 = 60° Radius = r Now, Area of minor segment = Area of segment APB = Area of sector OAPB – Area of ΔOAB Let’s find separately Area of sector OAPB Area of sector OAPB = 𝜃/360× 𝜋𝑟2 = 60/360 × 𝜋𝑟2 = 𝟏/𝟔 𝝅𝒓^𝟐 Area of Δ AOB In ∆ AOB, OA = OB = r ∠ AOB = 60° We know that Angles opposite equal side are equal ∴ ∠ OAB = ∠ OBA By Angle sum property ∠ OAB + ∠ AOB + ∠ OBA = 180° ∠ OAB + 60° + ∠ OBA = 180° Putting ∠ OAB = ∠ OBA ∠ OAB + 60° + ∠ OAB = 180° 2 × ∠ OAB + 60° = 180° 2 × ∠ OAB = 180° – 60° 2 × ∠ OAB = 120° ∠ OAB = (120°)/2 ∠ OAB = 60° Thus, ∠ OBA = ∠ OAB = 60° So, all 3 angles are 60° Which means ∆ OAB is an equilateral triangle We need to find area of equilateral triangle Area of equilateral triangle = √𝟑/𝟒 × 𝑺𝒊𝒅𝒆^𝟐 Putting = = √𝟑/𝟒 × 𝑺𝒊𝒅𝒆^𝟐 Area Δ AOB = 1/2 × Base × Height We draw OM ⊥ AB ∴ ∠ OMB = ∠ OMA = 90° And, by symmetry M is the mid-point of AB ∴ BM = AM = 1/2 AB Thus, ∠ OBA = ∠ OAB = 60° So, all 3 angles are 60° Which means ∆ OAB is an equilateral triangle And, Length of side = r We need to find area of equilateral triangle Area of equilateral triangle = √𝟑/𝟒 × 𝑺𝒊𝒅𝒆^𝟐 Putting Side = r = √𝟑/𝟒 𝒓^𝟐 Thus, Area of minor segment = Area of sector OAPB – Area of ΔOAB Let’s find separately Area of segment APB = Area of sector OAPB – Area of ΔOAB = 1/6 𝜋𝑟^2−√3/4 𝑟^2 = 𝒓^𝟐 (𝝅/𝟔−√𝟑/𝟒) Note: The question has a typo, we need to prove area is = 𝑟^2 (𝜋/6−√3/4) and not 𝜋𝑟^2 (1/6−√3/4)