In the diagram above, ADF is a right triangle. BCED is a square with

Question

square in a triangle.png In the diagram above, ADF is a right triangle. BCED is a square with an area of 12. What is the area of triangle ADF? Statement #1 : angle DCF = 75 degrees Statement #2: AB:EF = 3 Geometry is absolutely beautiful! This lovely problem is one of a set of ten DS practice problems on geometry. To see this others, as well as the OE to this question, see: GMAT Data Sufficiency Geometry Practice Questions Mike :-)

chetan2u · Answer

Hi,
since the area is given, we will know each side of square, so BD = DE =  \sqrt{12} = 2\sqrt{3} .......

lets see the statement.. 
  I : angle DCF = 75.... 
angle DCF = angle DCE + angle ECF... 
angle DCF is 75, and angle DCE will be 45 as the diagonal of square BISECTS the angle at vertex, that is 90/2 = 45....
75 = 45 + angle ECF .......angle ECF = 30....

CEF is 30-60-90 triangle...
opposite 60 is side CE, which is  2\sqrt{3} .... so EF = 2...
side DF = DE + EF =  2\sqrt{3}+2 ...

Now since angle DCF is 60.. ADF is also 30-60-90 triangle...
 we know one side DF, and thus we can find other sides and from tha the AREA

Suff

II : AB:EF = 3..... 
3 EF = AB.... 
let EF = x, so AB = 3x.....

You can calculate AREA of ADF in two ways and equate each other -

1)  area of ADF = area of CAD + Area of CDF  = \frac{1}{2} * AD *DF= \frac{1}{2} * ( 2\sqrt{3}+3x) * 2\sqrt{3} + \frac{1}{2} * ( 2\sqrt{3}+x) * 2\sqrt{3}

2)  area of ADF  =  \frac{1}{2}*AD*DF = \frac{1}{2} * ( 2\sqrt{3}+3x) *( 2\sqrt{3}+x) ..

Equate two and you will get x^2 = 4... x cannot be -ive so x= 2..
 No we know sides and we can find Area of ADF 
Suff

D

adiagr · Answer

Statement #1 : angle DCF = 75 degrees

We know that angle DCE = 45 degrees. (diagonal bisects the right angle; BCDE is a square).

Angle ECF = 75-45 = 30 degrees
Now because CEF = 90 degrees; Therefore Angle EFC = 60 degrees.

Also because ADF is a right angled triangle, Angle DAF = 90 - 60 = 30 degrees.

Now because I know angles; I can get EF and AB.

CEF is a 30-60-90 triangle. Its sides are in a ratio of 1: \sqrt{3}: 2. We know CE = 12. We can get EF. (No need to calculate. This is DS)
Once EF is known, DF (base of bigger triangle is known; DF = DE+EF)

Similarly, we can also get AB. ABC is also a 30-60-90 triangle. BC is parallel to DF and so angle BCA = angle EFC = 60 degrees.

Once AB is known, AD is also known.

Area of complete triangle is known. (0.5 AD * DF)

Sufficient

Statement #2:  AB:EF = 3

Say Ab = 3x, then EF will be x 
Triangles ABC and CEF are similar. (Angles are same.)

Therefore by triangle property:

\frac{AB}{CE}  =  \frac{BC}{EF}

\frac{3x}{12}  =  \frac{12}{x}

x is known.

Again Base of triangle and Height is known. Area can be calculated.

Sufficient.

D is the Answer.

CEdward · Answer

I got a lost here chetan

1) area of ADF = area of CAD + Area of CDF =12∗AD∗BF

Where is BF coming from?

chetan2u · Answer

It is actually DF as we are equating the two terms.

Typo, edited. Thank you

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In the diagram above, ADF is a right triangle. BCED is a square with

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In the diagram above, ADF is a right triangle. BCED is a square with

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