Question:hard

There are two concentric circles such that the area of the outer circle is four times the area of the inner circle. If A, B and C are three distinct points on the perimeter of the outer circle such that AB and AC are tangents of the inner circle, what is the area of the triangle ABC?

Statement (1): The area of the outer circle is 12 sq.cm

Statement (2): The area of the region between the two circles is 9 sq.cm

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Use R = 2r from the area condition, recognize the tangent configuration forces triangle ABC to be equilateral with circumradius R, then check if each statement alone gives you a numeric value for R (or the circle areas).

Updated On: Jul 20, 2026
  • If the data in statement (1) alone is sufficient to answer the question, but the data in statement (2) alone is not sufficient.
  • If the data in statement (2) alone is sufficient to answer the question, but the data in statement (1) alone is not sufficient.
  • If the data in both the statements together are needed to answer the question.
  • If either statement (1) alone or statement (2) alone is sufficient to answer the question.
  • If the data in neither statement (1) nor statement (2) is sufficient to answer the question, and more data is needed.
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The Correct Option is D

Solution and Explanation

Since the two circles are concentric with the outer area four times the inner, $R=2r$ where $R,r$ are the outer and inner radii. This particular ratio, $R=2r$, is the textbook relationship between a triangle's circumradius and inradius when the triangle is equilateral (this comes from Euler's formula $OI^2=R^2-2Rr$, which forces the incentre $I$ to coincide with the circumcentre $O$, i.e. concentric circles, only for an equilateral triangle). So once AB and AC are tangent to the inner circle and A, B, C lie on the outer circle with the two circles concentric, triangle ABC has to be equilateral, with the inner circle playing the role of its incircle and the outer circle its circumcircle.

For an equilateral triangle, Area $=\frac{3\sqrt3}{4}R^2$, so we just need a number for $R^2$, which is the same as knowing the outer circle's area, since $\text{Area}_{outer}=\pi R^2$.

Statement (1) hands us $\text{Area}_{outer}=12$ directly, so $R^2=12/\pi$ and the triangle's area is $\frac{3\sqrt3}{4}\cdot\frac{12}{\pi}=\frac{9\sqrt3}{\pi}$ sq.cm — a fixed number, sufficient alone.

Statement (2) gives the annulus area (outer minus inner) as $9$. Using $\text{Area}_{outer}=4\times\text{Area}_{inner}$, if inner $=y$ then $4y-y=9 \Rightarrow y=3$, so outer $=12$ — the same value as statement (1) produces, leading to the same triangle area — sufficient alone as well.

Both statements independently resolve the outer circle's area (and therefore the triangle's area), so either one alone answers the question: option (4).

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