Equilateral triangle BDF Is Inscribed in Equilateral Triangle ACE GMAT Data Sufficiency

Question: Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE?

(1) Angle DFE is 90º
(2) The length of AF is 10*sqrt(3)

1. Statement (1) ALONE is sufficient but statement (2) ALONE is not sufficient.
2. Statement (2) ALONE is sufficient but statement (1) ALONE is not sufficient.
3. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
4. EACH statement ALONE is sufficient.
5. Statements (1) and (2) TOGETHER are not sufficient.

‘Equilateral triangle BDF is inscribed in equilateral triangle ACE, as shown in the figure above. The shaded region is what fraction of the area of the triangle ACE’- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Quantitative Review". GMAT Quant section consists of a total of 31 questions. GMAT Data Sufficiency questions consist of a problem statement followed by two factual statements. GMAT data sufficiency comprises 15 questions which are two-fifths of the total 31 GMAT quant questions.

Solution and Explanation:

Approach Solution 1:

This question can be solved by only one approach.
It is given in the question that there is an equilateral triangle BDF inscribed in the equilateral triangle ACE. It is asked in the question that shaded region is what part of the total area of triangle ACE.
For an equilateral triangle all angles are equal to 60 degrees and the area is
A = \(a^2\)\(\sqrt{3}\)/4 where a is the length of the side.
In statement 1,
It is given that angle DFE is 90 degrees
Angle ABF and BDC has to be equal to 90 degrees.
Proof:
BDC = 180 - BDF - EDF = 180 - 60 - 30 = 90
Same for ABF
Also it is given that,
Angle DEF, BCD, BAF = 60 degree, then triangle DFE, BCD, BAF are 30-60-90 degree angles.
In these right angled triangle sides are in the ratio of 1:\(\sqrt{3}\):2
So if DE = 2x (hypotenuse in the right angled triangle DFE), then DC = x (smaller side in right angled triangle BCD) and BD = \(\sqrt{3}\)x ( larger side in the triangle BCD, also the side of inscribed triangle)
So the side of triangle ACE would be CE = DC + DE = x + 2x = 3x
Area of the colored region = area of triangle BDC = BD * DC /2 = \(\sqrt{3}\)x*x/2 = \(\sqrt{3}\)\(x^2\)/2 and the area of the equilateral triangle ACE = \(a^2\)\(\sqrt{3}\)/4 = \((3x)^2\)*\(\sqrt{3}\)/4 = 9\(x^2\)\(\sqrt{3}\)/4
Area of triangle BDC / area of the triangle ACE = \(\sqrt{3}\)\(x^2\)/2 * 4/(9\(x^2\)*\(\sqrt{3}\)) = 2/9
This is sufficient to get the answer.
In statement 2,
The length of AF is 10\(\sqrt{3}\). Multiple breakdowns are possible and therefore multiple area ratios can be found.
This is not sufficient to get the answer.
Therefore the correct option is option A

Correct Answer: A

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