In-radius of equilateral triangle of side a =
Diameter of larger circle =
Let us say common tangent PQ touches the two circle at R, center of smaller circle is I.
Now, PQ is parallel to BC. AR is perpendicular to PQ. Triangle PQR is also an equilateral triangle and AORID is a straight line. (Try to establish each of these observations. Just to maintain the rigour.)
AD = a
RD =
AR =
=
AR = AD.
Radius of smaller circle = radius of larger circle
Radius of smaller circle =
Area of smaller circle = πr^{2}
π =
Area of △ = a^{2}
Ratio = : a^{2}
π : 27. Answer choice (c)
Correct Answer: π : 27
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