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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.

Find the area of the minor segment of a circle of radius 42 cm,

Question 20 - CBSE Class 10 Sample Paper for 2019 Boards - Part 2
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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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.