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Question 6

In the figure given, the radius of the circle is 15 cm. The angle subtended by the chord AB at the centre O is 60. Find the area of the major and minor segments.

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Solution

Radius of the circle = 15 cm
ΔAOB is an isosceles triangle as two sides are equal.
A=B
Now, sum of all angles of triangle =180
A+B+O=180
2A=18060
A=1202
A=60

Hence, the triangle is equilateral as A=B=C=60
OA=OB=AB=15 cm
Area of equilateral ΔAOB=34×(OA)2
=34×152
=(22534) cm2=97.3 cm2

Angle subtended at the centre by minor segment =60
Area of the sector making angle θ
=(θ360)×πr2
Area of minor sector making angle 60
=(60360)×πr2 cm2
= (16)×152π cm2=2256π cm2
=(2256)×3.14 cm2=117.75 cm2

Area of the minor segment
= Area of minor sector - Area of equilateral ΔAOB
=117.75 cm297.3 cm2=20.4 cm2

Angle made by major sector =36060=300
Area of the sector making angle 300
=(300360)×πr2 cm2
=(56)×152π cm2=11256π cm2
=(11256)×3.14cm2=588.75 cm2

Area of major segment
= Area of major sector + Area of equilateral ΔAOB
=588.75 cm2+97.3 cm2=686.05 cm2

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