If the length of the side of the inscribed square measures 12 inches

Question

https://gmatclub.com/forum/download/file.php?id=64005 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 Project DS Butler Data Sufficiency (DS3) For DS butler Questions Click Here 1.PNG

asishron29181 · Answer

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
Consider the smaller triangles at the two corners.
its a 30-60-90 triangle.
Since corresponding to 60 is 12
√3 x = 12
x = 4√3
2x = 8√3
From this we can find the area of two smaller triangles at the corner but we still don't know anything about the top triangle.
Insufficient.

(2) The height o the triangle ABC from point C is 6√3+12
Clearly we can say that height of top triangle is 6√3
But since we don't have any idea about corner triangles hence insufficient again.

Combining both we can figure out areas of two corner and one top triangle, hence sufficient to figure out area of ABC.

IMO C

Shahalikhan · Answer

IMO D

(1) ABC is an equilateral triangle

all angles are 60 degree. we can find AB and then find area sufficient

(2)by triangle proportionality theorem

a line parallel to one side creates similar triangles

we can find the ratio of sides 
thus ratio of areas and then area

sufficient

Bunuel · Answer

This question is a part of  Are You Up For the Challenge: 700 Level Questions  collection.

andrenoble · Answer

Would like to add several points for clarification:

Statement I

https://gmatclub.com/forum/download/file.php?id=64005

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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If the length of the side of the inscribed square measures 12 inches

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