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∘ ⇒2∠A=180∘−60∘ ⇒∠A=120∘2 ⇒∠A=60∘
Hence, the 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 the sector making angle θ =(θ360∘)×πr2 Area of minor sector making angle 60∘ =(60∘360∘)×πr2cm2 = (16)×152πcm2=2256πcm2 =(2256)×3.14cm2=117.75cm2
Area of the minor segment = Area of minor sector - Area of equilateral ΔAOB =117.75cm2−97.3cm2=20.4cm2
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