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Question
Find the area of the segment shown in Fig. 2, if the radius of the circle is 21 cm and ∠AOB = 120° `( "Use" π = (22)/(7))`
Solution
Construction: Draw a line passing through O and perpendicular to AB.
In ΔAOM and ΔBOM,
∠AMO = ∠BMO (by construction)
AO = BO (radius of the same circle)
OM = OM ( common side)
∴ ΔAOM ≅ Δ BOM (By RHS congruence rule)
We have , ∠AMO = ∠BMO = 60° (By CPCT) ....(i)
AM = BM (By CPCT) .......(ii)
In ΔAOM,
sin 60° = `"AM"/"OA" = "AM"/(21)`
⇒ `"AM"/(21) = sqrt3/(2)`
⇒ AM = `(21sqrt3)/(2) "cm"`
Also, cos 60° = `"OM"/"OA"`
⇒ `"OM"/"OA" = (1)/(2)`
⇒ `"OM"/"OA" = (21)/(2) "cm"`
AB = AM + MB = 2AM = `21sqrt3 "cm"` .......[from (ii)]
Area of sector AOB = `(120)/(360) · πr^2 = (1)/(3)· (22)/(7)· 21^2 = 462"cm"^2`
Area of ΔAOB = `(1)/(2) xx "OM" xx "AB" = (1)/(2) xx (21)/(2) xx 21 sqrt3 = (444sqrt3)/(4) "cm"^2 ≈ 191 "cm"^2`
Required area of segment = Area of sector AOB -">− Area of ΔAOB
= 462 - 191 = 271 cm2 (approx.)
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