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Home/ Questions/Q 1567
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Atiksh Tomar
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Atiksh Tomar
Asked: October 29, 20202020-10-29T05:30:08+00:00 2020-10-29T05:30:08+00:00In: Class 10

A chord of a circle of radius 15 cm subtends an angle of 60° at the Centre. Find the areas of the corresponding A chord of a circle of radius 15 cm subtends an angle of 60° at the Centre. Find the areas of the corresponding minor and major segments of the circle. (Use π= 3.14 and √3 = 1.73)minor and major segments of the circle. (Use π= 3.14 and √3 = 1.73)

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NCERT Solutions for Class 10 Science Chapter 12
Important NCERT Questions
Areas Related To Circles Class,
NCERT Books for Session 2022-2023
CBSE Board and UP Board Others state Board
EXERCISE 12.2
Page No:230
Questions No:6

2020-2021areas related to circles classcbsechapter 12class10mathematicsncert
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    1. Gajasura
      2021-07-31T11:01:20+00:00Added an answer on July 31, 2021 at 11:01 am

      Get Hindi Medium and English Medium NCERT Solution for Class 10 Maths to download.
      Please follow the link to visit website for first and second term exams solutions.
      https://www.tiwariacademy.com/ncert-solutions/class-10/maths/chapter-12/

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      pappu07
      2023-02-21T03:38:56+00:00Added an answer on February 21, 2023 at 3:38 am
      A chord of a circle of radius 15 cm subtends an angle of 60° at the Centre. Find the areas of the corresponding A chord of a circle of radius 15 cm subtends an angle of 60° at the Centre. Find the areas of the corresponding minor and major segments of the circle. (Use π= 3.14 and √3 = 1.73)minor and major segments of the circle. (Use π= 3.14 and √3 = 1.73)

      Radius of circle 15 cm
      Area of sector OPRQ
      = 60°/360° × πr² = 1/6 × π(15)² = 1/6 × 3.14 × 15 × 15 = 117.75 cm²
      Area of minor segment
      = Area of minor sector OPRQ – Area of △0PQ
      = 117.75 cm² – √3/4(15) cm² [As triangle OPQ is an equilateral triangle]
      = 117.75 cm² – 225√3/4 cm²
      = 117.75 cm² – 56.25 x 1.73 cm²
      = 231 cm² – 97.3125 cm²
      = 20.4375 cm²
      Area of major segment = Area of circle – Area of minor segment
      = πr² – 20.4375 cm²
      = π(15)² – 20.4375] cm²
      = [3.14 x 15 x 15 – 20.4375] cm²
      = [706.5 – 20.4375] cm²
      = 686.0625 cm²

      For better understanding to the above question see here😎👇

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