Question 20

Find the area of the minor segment of a circle of radius 42 cm, if length of the corresponding arc is 44 cm.       1. Class 10
2. Solutions of Sample Papers for Class 10 Boards
3. CBSE Class 10 Sample Paper for 2019 Boards

Transcript

Question 20 Find the area of the minor segment of a circle of radius 42cm, if length of the corresponding arc is 44cm. Let’s first draw the figure We first have to find the angle We know that Length of arc = 𝜃/360 × 2πr Putting values 44 = 𝜃/360 × 2 × 22/7 × 42 44 = 𝜃/180 × 22/7 × 42 44 = 𝜃/180 × 22 × 6 (44 × 180)/(22 × 6) = θ 2 × 30 = θ 60 = θ θ = 60° Now, we have to find Area of segment Area of segment APB = Area of sector OAPB – Area of ΔOAB Area of sector OAPB Area of sector OAPB = 𝜃/360×𝜋𝑟2 = 60/360×22/7×42×42 = 1/6×22/7×42×42 = 1/6×22×6×42 = 22×42 = 924 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 (Both 90°) OA = OB (Both radius) OM = OM (Common) ∴ Δ OMA ≅ Δ OMB (By R.H.S congruency) ⇒ ∠ AOM = ∠ BOM (CPCT) ∴ ∠ AOM = ∠ BOM = 1/2 ∠ BOA ⇒ ∠ AOM = ∠ BOM = 1/2 × 60° = 30° Also, since Δ OMB ≅ Δ OMA (CPCT) …(1) ∴ BM = AM ⇒ BM = AM = 1/2 AB In right triangle Δ OMA sin O = (Side opposite to angle O)/Hypotenuse sin 30° = AM/AO 1/2=𝐴𝑀/42 AM = 42/2 = 21 In right triangle Δ OMA cos O = (𝑆𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒 cos 30° = 𝑂𝑀/𝐴𝑂 √3/2=𝑂𝑀/21 OM = √3/2 × 42 = 21√3 From (1) AB = 2AM Putting value of AM AB = 2 × 21 AB = 42 cm Now, Area of Δ AOB = 1/2 × Base × Height = 1/2 × AB × OM = 1/2 × 42 × 21√3 = 441√3 cm2 Now, Area of segment APB = Area of sector OAPB – Area of ΔOAB = (924 – 441√3) cm2

CBSE Class 10 Sample Paper for 2019 Boards

Class 10
Solutions of Sample Papers for Class 10 Boards 