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.

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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 = πœƒ/360 Γ— 2Ο€r Putting values 44 = πœƒ/360 Γ— 2 Γ— 22/7 Γ— 42 44 = πœƒ/180 Γ— 22/7 Γ— 42 44 = πœƒ/180 Γ— 22 Γ— 6 (44 Γ— 180)/(22 Γ— 6) = ΞΈ 2 Γ— 30 = ΞΈ 60 = ΞΈ ΞΈ = 60Β° Now, we have to find Area of segment Area of segment APB = Area of sector OAPB – Area of Ξ”OAB Area of sector OAPB Area of sector OAPB = πœƒ/360Γ—πœ‹π‘Ÿ2 = 60/360Γ—22/7Γ—42Γ—42 = 1/6Γ—22/7Γ—42Γ—42 = 1/6Γ—22Γ—6Γ—42 = 22Γ—42 = 924 cm2 Finding area of Ξ” AOB Area Ξ” AOB = 1/2 Γ— Base Γ— Height We draw OM βŠ₯ AB ∴ ∠ OMB = ∠ OMA = 90Β° In Ξ” OMA & Ξ” OMB ∠ OMA = ∠ OMB (Both 90Β°) OA = OB (Both radius) OM = OM (Common) ∴ Ξ” OMA β‰… Ξ” OMB (By R.H.S congruency) β‡’ ∠ AOM = ∠ BOM (CPCT) ∴ ∠ AOM = ∠ BOM = 1/2 ∠ BOA β‡’ ∠ AOM = ∠ BOM = 1/2 Γ— 60Β° = 30Β° Also, since Ξ” OMB β‰… Ξ” OMA (CPCT) …(1) ∴ BM = AM β‡’ BM = AM = 1/2 AB In right triangle Ξ” OMA sin O = (Side opposite to angle O)/Hypotenuse sin 30Β° = AM/AO 1/2=𝐴𝑀/42 AM = 42/2 = 21 In right triangle Ξ” OMA cos O = (𝑆𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œ π‘Žπ‘›π‘”π‘™π‘’ 𝑂)/π»π‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘ π‘’ cos 30Β° = 𝑂𝑀/𝐴𝑂 √3/2=𝑂𝑀/21 OM = √3/2 Γ— 42 = 21√3 From (1) AB = 2AM Putting value of AM AB = 2 Γ— 21 AB = 42 cm Now, Area of Ξ” AOB = 1/2 Γ— Base Γ— Height = 1/2 Γ— AB Γ— OM = 1/2 Γ— 42 Γ— 21√3 = 441√3 cm2 Now, Area of segment APB = Area of sector OAPB – Area of Ξ”OAB = (924 – 441√3) cm2

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.