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radius of inscribed circle r = a* √3/6
a given = 24 so radius ; 4*√3 area of circle pi * 48
and area of equilateral ∆ = √3/4 * side ^2 side= 24 so area = 144√3
shaded region area = \(144√3 - 48π\) cm^2
OPTION C


Bunuel

A circle is inscribed in an equilateral triangle of side 24 cm, as shown above. What is the area of the shaded region?

A. \(148√3 - 36π\) cm^2

B. \(144√3 - 36π\) cm^2

C. \(144√3 - 48π\) cm^2

D. \(121√3 - 48π\) cm^2

E. \(121√3 - 36π\) cm^2


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area of equilateral ∆ABC = \(\frac{\sqrt{3}}{4}*side^2 = 144 \sqrt{3}\)
And inradius = \(\frac{1}{3}*12\sqrt{3} = 4 \sqrt{3}\)
So required area = ∆ABC - area of incircle
=\(144\sqrt{ 3} - 48 \pi\)
Answer: C
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IMO C
Concepts and Formulae used:
Based on the question: Area of Shaded Region = [Equilateral ∆ - Area of Circle]
Concept: Radius of circle inscribed in an equilateral triangle = (1/3)* Height of equilateral triangle = (1/3)*(√3/2)*Side

Now, the steps:
Step 1 : Area of equilateral ∆ =(√3/4)∗242=144√3=(√3/4)∗242=144√3
Step 2: Area of Circle =πr*2=πr*2
Step 3: Radius of Circle =(1/3)*(√3/2)*24=4√3=(1/3)*(√3/2)*24=4√3
Step 4: Area of Circle =π* (4√3)*2=π*48=π*(4√3)2=48π
Answer: Shaded Area =[144√3−48π]
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