Question

formula to find minor segment

Answer 1

(πr²theta/360)-(r²sintheta /2)
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Answer 2

The area of the minor segment when angle θθ and radius rr are given:

Area of segment == area of sector AOBCAOBC ±± area of ΔAOBΔAOB
=12r2θ±12r2sinθ=12r2θ±12r2sin⁡θ
=12r2(θ−sinθ)=12r2(θ−sin⁡θ)




Now the area of the major segment == area of circle −− area of the minor segment
=12r2(2π−θ+sinθ)=12r2(2π−θ+sin⁡θ)

 

Example:

A chord ABAB of a circle of radius 1515cm makes an angle of 60∘60∘at the center of the circle. Find the area of the major and minor segment.

 

Solution:




Given that ∠AOB=60∘∠AOB=60∘, radius, r=15r=15cm

∴∴ area of the sector OABOAB =θ360×πr2=θ360×πr2
=60360×3.1415×15×15=117.75=60360×3.1415×15×15=117.75 square cm
∴∴ area of ΔOABΔOAB =12r2sinθ=12r2sin⁡θ
=12×15×15×sin60∘=225×1.734=97.31=12×15×15×sin⁡60∘=225×1.734=97.31 square cm
Area of the minor segment == area of sector OABOAB−− area of  ΔOABΔOAB
=117.75−97.31=20.44=117.75−97.31=20.44 square cm
Area of the circle =πr2=πr2
=3.1415×(15)2=3.1415×225=706.5=3.1415×(15)2=3.1415×225=706.5 square cm
Area of the major segment == area of the circle −− area of the minor segment
=706.5−20.4=686.1=706.5−20.4=686.1 square cm.



Read more: https://www.emathzone.com/tutorials/geometry/area-of-a-segment.html#ixzz5VhhZ8XCf