Ex 11.1, 4 - Chapter 11 Class 10 Areas related to Circles

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 11.1, 4
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding :
minor segment (Use π= 3.14)
Given that
OA = OB = radius = 10 cm
θ=90°
Now,
Area of segment APB = Area of sector OAPB – Area of ΔAOB
Area of sector OAPB
Area of sector OAPB = θ/(360°)× πr2
= 90/360 × 3.14 × (10)2
= 𝟏/𝟒 × 𝟑.𝟏𝟒 × 𝟏𝟎𝟎
= 1/4× 314
= 78.5 cm2
Area of ΔAOB
Now, ΔAOB is a right triangle, where ∠ O = 90°
having Base = OA & Height = OB
Area of Δ AOB = 1/2 × Base × Height
= 𝟏/𝟐 × OA × OB
= 1/2 × 10 × 10
= 50 cm2
Now,
Area of segment APB = Area of sector OAPB – Area of ΔAOB
= 78.5 – 50
= 28.5 cm2
Ex 11.1, 4 (Method 1)
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding :
(ii) major sector (Use π= 3.14)
Here, θ = 90°
& radius = r = 10 cm
Area of major sector = ((𝟑𝟔𝟎° − 𝜽))/(𝟑𝟔𝟎°)×𝛑𝐫𝟐
= ((360 − 90))/360 × 3.14 × (10)2
= 𝟐𝟕𝟎/𝟑𝟔𝟎 × 𝟑.𝟏𝟒 × 𝟏𝟎𝟎
= 3/4× 314
= 235.5 cm2
Ex 11.1, 4(Method 2)
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding :
(ii) major sector (Use π= 3.14)
Area of major sector
= Area of circle – Area of sector OAPB
Area of circle = πr2
= 3.14 × (10)2
= 3.14 × 100
= 314 cm2
Now,
Area of major sector = Area of circle – Area of sector OAPB
= 314 – 78.5
= 235.5 cm2

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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