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Question
If area of a circle inscribed in an equilateral triangle is 48π square units, then perimeter of the triangle is
Options
17 \[\sqrt{3}\]units
36 units
72 units
48\[\sqrt{3}\]units
Solution
Let the circle of radius r be inscribed in an equilateral triangle of side a.
Area of the circle is given as 48π.
`⇒ pir^2=48pi`
`⇒ r^2=48`
`⇒ r=4sqrt3`
Now, it is clear that ON⊥BC. So, ON is the height of ΔOBC corresponding to BC.
Area of ΔABC = Area of ΔOBC + Area of ΔOCA + Area of ΔOAB = 3 × Area of ΔOBC
`sqrt3/4xxa^2=3xx1/2xxBCxxON`
`sqrt3/4xxa^2=3xx1/2xxaxxr`
`sqrt3/4xxa^2=3xx1/2xxaxx4sqrt3`
`a=24`
Thus, perimeter of the equilateral triangle = 3 × 24 units = 72 units
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