Question

In the given fig. ABC is an equilateral triangle inscribed in a
circle of radius 4 cm and centre 0. Show that the area of the
shaded region is 4/3(4π-3√3) cm2

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Answer 1

Given :-

• ABC is a Equaliteral ∆.
• O is centre .
• Radius of circle = 4cm.

To prove :-

• Area of shaded region = 4/3(4π-3√3) cm².

Concept & Formula used :-

• Angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle .
• Area of ∆ = (1/2) * Side1 * side2 * (Angle b/w both sides).
• Area of sector with angle @ = (@/360°) * π * (radius)²

Solution :-

→ ∠BAC = 60° (Equaliteral ∆).

→ ∠BOC = 120° (Double).

→ BO = OC = Radius = 4cm.

Hence,

→ Area of sector BOC = (120°/360°) [ π * (4)² ]

→ Area of sector BOC = (1/3) * (16π) cm² ------- Equation (1).

__________________

Now,

In ∆BOC we have :-

→ ∠BOC = 120° (Double).

→ BO = OC = Radius = 4cm.

Hence,

→ Area ∆BOC = (1/2) * BO * OC * (sin∠BOC)

→ Area ∆BOC = (1/2) * 4 * 4 * sin120°

→ Area ∆BOC = 8 * (√3/2)

→ Area ∆BOC = 4√3 cm² . --------- Equation (2).

__________________

Therefore,

→ Required Shaded Area = Area of sector - Area of ∆BOC

→ Required Shaded Area = Equation (1) - Equation (2)

→ Required Shaded Area = [ (1/3) * (16π) ] - 4√3

→ Required Shaded Area = (4/3) [ 4π - 3√3 ] cm² (Ans.)

Hence, Area of shaded Region will be (4/3) [ 4π - 3√3 ] cm² .