CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A chord of a circle of radius 12cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use π=3.14and 3=1.73)


Open in App
Solution

Step 1: Find the area of minor sector of the circle

Radius, r=12cm

Draw a perpendicular ODon chord ABand it will bisect chord AB.

So, AD=DB

Radius =r=12cm

Area of triangle =12×base×height

Area of segment=Area of the sector -the area of a triangle

The area of the minor sector =θ360×πr2

=120°360°×3.14×122

=150.72cm2

Step 2: Find the area of ΔAOB

From the ΔAOB,

OAB=180°-(120°)2=30°OA=OB=r,andintriangletheangleoppositetoequalsidesareequal

Now, Cos30°=ADOA

=32=AD12

Or, AD=63cm

We know that ODbisects AB.

So,

AB=2×AD=123cm

Now, Sin30°=ODOA

=12=OD12

OD=6cm

So, the area of ΔAOB=12×base×height

Here, base AB=123and

Height=OD=6

So, area of ΔAOB=12×123×6=363cm=62.28cm2

Step 3: Find the area of Minor segment

AreaofthecorrespondingMinorsegment=AreaoftheMinorsectorAreaofΔAOB

=150.72cm262.28cm2

=88.44cm2

Step 4: Find the area of major segment

Area of major segment =Area of circle -Area of minor segment

=πr2-88.44=3.14×122-88.44=452.16-88.44=363.72cm2

Hence, the area of the corresponding segment of the circle are 88.44cm2 and 363.72cm2


flag
Suggest Corrections
thumbs-up
3
BNAT
mid-banner-image