Equilateral triangle BDF is inscribed in equilateral triangl

As you said, easier to solve, many thanks!!!!!

Good solution in steps given.

Bumping for review and further discussion.

GEOMETRY: Shaded Region Problems!

\(\angle {BDC}=180-\angle {BCD}-\angle{EDF}=180-60-30\) -

Could you explain the logic for taking EDF?

Since you asked for further discussion, here goes.

I more intuited the answer than solved. I know that each interior angle of both equilateral triangles is 60. If I am told that DFE = 90, I can figure out that the other 3 triangles are 30-60-90 and congruent (Each 60-degree angle is across from one side of the smaller equilateral triangle; thus all corresponding sides for the right triangles are the same "x(sqrt3)"). Because I know the relative dimensions of triangles BCD, ABF, DFE, and BDF, I can figure out the relative areas of the triangles. For example, I know I can make BD sqrt3, BC 2, and CD 1. I can figure out the areas of all 4 smaller triangles, add these areas to get ACE's area, and thus determine the fraction of ACE that BCD represents. I didn't actually bother to assign values or do the calculations because it was clear Statement 1 is sufficient.

Eliminate B, C, and E.

Statement 2 only tells me AF. It doesn't provide me with any information about any other sides or angles in the figure. Thus, I would not be able to solve for area. The information is useless. Insufficient.

Eliminate D. The answer is A.

It should be \(\angle {BDC}=180-\angle {BDF}-\angle{EDF}=180-60-30\) (those 3 angles make up a straight line, which is 180 degrees).

Hope it's clear.

Statement I

I can explain why without solving it I think

Knowing that that angle is 90 degrees makes this suff because you are told both the triangle and the non shaded inner circle are equilateral - meaning you have a decent amount of this filled out for you in terms of angles. Fill the angles out and you'll notice that you end up with the shaded area being a 30 60 90 triangle, and the right angle you are given is a 30 60 90 triangle, that the remaining not given triangle is a 30 60 90 triangle and ofc you are already given the inner triangle is a 60 60 60 triangle on top of the entire triangle being a 60 60 60 triangle.

Okay well 30 60 90 triangles are special in that the lengths are x, xrt3, and 2x - right? well this is a fraction problem so it doesn't matter what the lengths actually are - all that matters is you know the proportion of the shaded area in relation to the whole triangle. Considering you now the bare min. lengths of every side of every inner triangle and that the sides of the whole must add up to thethe same lengths relative to these smaller triangles lengths summed up then you can find that fractional amount of that portion to the entire thing.

I hope that makes sense if the math of it all seems too daunting. IMO it'd be a huge waste of time to actually do this out on test day - I find these so much faster to look at and use logic and just see the simple angle relationships instead of losing it on how to actually prove it with math formulas.

Statement II
This is not sufficient because even if you know that length you can still conceive of a situation where the inner triangle is shifting onto different points of the outer triangle points and thus the length of AE (and thus AC and CE) could have a number of lengths with statement II STILL being true - meaning the angle relationships are shifting with that triangle. The shifting of that inner 60 60 60 triangle would thus cause the size of the shaded area to change with it and thus make that fraction change.

Base line is that the additional angle allows you to follow a string of logic to validate every single angle in the entire thing whereas the one given length is just not in relation to anything.