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

Introducing your new favourite teacher - Teachoo Black, at only β‚Ή83 per month


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

Davneet Singh's photo - Teacher, Engineer, Marketer

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.