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Bunuel
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Bunuel

If the length of the side of the inscribed square measures 12 inches, what is the area of triangle ABC?

(1) ABC is an equilateral triangle
(2) The height o the triangle ABC from point C is \(6\sqrt{3}+12\)


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Would like to add several points for clarification:

Statement I



Since the triangle ABC is equilateral, it gives us an equilateral triangle with "on top" of the square (angle C = 60 degrees, the triangle is isosceles since square has parallel sides, hence the triangle is equilateral), and we can calculate its area.

We also have 2 "30-60-90" triangles to the sides of the square, and we could calculate their area because we know that sides of such triangles have ratio of 1: \sqrt{3}.

We could sum all areas to get the area of the ABC.

Statement II

Since we have similar triangles, we know that the height of the triangle "on top" of the square will be 6\sqrt{3} (substract the height of the square from given). Through that it's possible to find that the triangle is actually equilateral (6^2 + (6\sqrt{3})^2 = 12^2), and after that you could continue the logic used above for Statement I.

Hence, both are independently sufficient -> D
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