Check sibling questions

Ex 12.3, 6 - In a circular table cover of radius 32 cm - Ex 12.3

Ex 12.3, 6 - Chapter 12 Class 10 Areas related to Circles - Part 2
Ex 12.3, 6 - Chapter 12 Class 10 Areas related to Circles - Part 3 Ex 12.3, 6 - Chapter 12 Class 10 Areas related to Circles - Part 4 Ex 12.3, 6 - Chapter 12 Class 10 Areas related to Circles - Part 5 Ex 12.3, 6 - Chapter 12 Class 10 Areas related to Circles - Part 6 Ex 12.3, 6 - Chapter 12 Class 10 Areas related to Circles - Part 7

This video is only available for Teachoo black users

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!


Transcript

Ex 12.3, 6 In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region). Area of design = Area of circle – area of triangle ABC For circle Radius = r = 32 cm Area of circle = πœ‹π‘Ÿ2 = 22/7Γ—322 = 22/7Γ—32Γ—32 = 22528/7 cm2 Area Ξ” ABC Now for equilateral triangle ABC, We find side first Let centre of circle be O. Join OA, OB and OC For Ξ” AOB, Ξ” BOC & Ξ” AOC OA = OB = OC OB = OC = OA AB = BC = AC Hence, AOB β‰… BOC β‰… AOC ∴ ∠"AOB"=∠"BOC"=∠AOC But, sum of angles around a point = 360Β° β‡’ ∠ AOB + ∠ BOC + ∠ AOC = 360Β° ∠ BOC + ∠ BOC + ∠ BOC = 360 3∠ BOC = 360Β° ∠ BOC = (360Β°)/3 ∠ BOC = 120Β° Also, Since Ξ” AOB, Ξ” AOC & Ξ” BOC are congruent Their areas would also be equal. Hence, Area Ξ” AOB = Area Ξ” AOC = Area Ξ” BOC So, Area of βˆ† 𝐡𝑂𝐢 = 1/3 (Area of ABC) Now, we find area Ξ” BOC. Finding area of Ξ” BOC Area Ξ” BOC = 1/2 Γ— Base Γ— Height We draw OM βŠ₯ BC ∴ ∠ OMB = ∠ OMC = 90Β° In Ξ” OMB & Ξ” OMC ∠ OMB = ∠ OMC OB = OC OM = OM ∴ Ξ” OMB β‰… Ξ” OMC β‡’ ∠ BOM = ∠ COM ∴ ∠ BOM = ∠ COM = 1/2 ∠ BOC β‡’ ∠ BOM = ∠ COM = 1/2 Γ— 120Β° = 60Β° Also, since Ξ” OMB β‰… Ξ” OMC ∴ BM = CM β‡’ BM = CM = 1/2 BC From (1) BM = 1/2BC 2BM = BC BC = 2BM Putting value of BM BC = 2 Γ— 16√3 BC = 32√3 Now, Area of Ξ” BOC = 1/2 Γ— Base Γ— Height = 1/2 Γ— BC Γ— OM = 1/2 Γ— 32√3 Γ— 16 = 16√3 Γ— 16 = 256√3 We know that Area of βˆ† BOC" = " 1/3 "Area of" βˆ† ABC 256 √3 = 1/3 Area of ABC 256 √3Γ—3 = Area of ABC 768 √3 = Area of ABC Area of βˆ† ABC=768√3 cm2 Now , Area of design = Area of circle – area of triangle ABC = 22528/7βˆ’768√3 cm2

Ask a doubt (live)
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.