Question 6
In the figure given, the radius of the circle is 15cm. The angle subtended by the chord AB at the centre O is 60∘. Find the area of the major and minor segments.
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∘
⇒2∠A=180∘−60∘
⇒∠A=120∘2
⇒∠A=60∘
Triangle is equilateral as ∠A=∠B=∠C=60∘
∴OA=OB=AB=15cm
Area of equilateral ΔAOB=√34×(OA)2
=√34×152
=(225√34)cm2=97.3cm2
Angle subtended at the centre by minor segment =60∘
Area of minor sector making angle 60∘
=(60∘360∘)×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 cm2−97.3 cm2=20.4 cm2
Angle made by major sector =360∘−60∘=300∘
Area of the sector making angle 300∘
=(300∘360∘)×πr2cm2
=(56)×152πcm2=11256πcm2
=(11256)×3.14cm2=588.75cm2
Area of major segment
= Area of major sector + Area of equilateral ΔAOB
=588.75cm2+97.3cm2=686.05cm2