A circle of radius unity is inscribed in an isosceles triangle. The least perimeter of the triangle is

6 \sqrt{3}

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9

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2 \sqrt{3}

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3 \sqrt{3}

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Solution

The correct option is B $6 \sqrt{3}$
The least perimeter of an isosceles triangle in which a circle of radius $r$ can be inscribed is $6 r \sqrt{3}$ .
In this case $= 1$ ,
$∴$ Required perimeter $= 6 \sqrt{3}$

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