Ex 11.1, 6 - Chapter 11 Class 10 Areas related to Circles

Last updated at April 16, 2024 by Teachoo

Transcript

Ex 11.1, 6
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)
In a given circle,
Radius (r) = 15 cm
And, 𝜽 = 60°
Now,
Area of segment APB = Area of sector OAPB – Area of ΔOAB
Finding Area of sector OAPB
Area of sector OAPB = 𝜃/360 × 𝜋𝑟2
= 60/360 × 3.14 × (15)2
= 1/6 × 3.14 × 15 × 15
= 1/2 × 3.14 × 5 × 15
= 117.75 cm2
Finding area of Δ AOB
We draw OM ⊥ AB
∴ ∠ OMB = ∠ OMA = 90°
And, by symmetry
M is the mid-point of AB
∴ BM = AM = 1/2 AB
In right triangle Δ OMA
sin O = (side opposite to angle O)/Hypotenuse
sin 𝟑𝟎° = 𝐀𝑴/𝑨𝑶
1/2=𝐴𝑀/15
15/2 = AM
AM = 𝟏𝟓/𝟐
In right triangle Δ OMA
cos O = (𝑠𝑖𝑑𝑒 𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡 𝑡𝑜 𝑎𝑛𝑔𝑙𝑒 𝑂)/𝐻𝑦𝑝𝑜𝑡𝑒𝑛𝑢𝑠𝑒
cos 𝟑𝟎° = 𝑶𝑴/𝑨𝑶
√3/2=𝑂𝑀/21
√3/2 × 15 = OM
OM = √𝟑/𝟐 × 15
From (1)
AM = 𝟏/𝟐AB
2AM = AB
AB = 2AM
Putting value of AM
AB = 2 × 1/2 × 15
AB = 15 cm
Now,
Area of Δ AOB = 1/2 × Base × Height
= 𝟏/𝟐 × AB × OM
= 1/2 × 15 × √3/2 × 15
= 1/2 × 15 × 1.73/2 × 15
= 97.3125 cm2
Therefore,
Area of segment APB
= Area of sector OAPB – Area of ΔOAB
= 117.75 – 97.3125
= 20.4375 cm2
Thus, Area of minor segment = 20.4375 cm2
Now,
Area of major segment = Area of circle – Area of minor segment
= πr2− 20.4375
= 3.14 × 𝟏𝟓𝟐− 20.4375
= 3.14 × 15 × 15−"20.4375"
= 706.5 – 20.4375
= 686.0625 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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