A sheet of paper ABDE is a 12-by-18-inch rectangle, as shown in Figure

Fix the first picture by drawing the lines which correspond to the way we folded our paper.
\(AF = FD, AC = CD\) so angle \(CAD = CDA\) and \(DAF = ADF => FDE = CAB => DFE = BCD => AFD = ACD => CAD = DAF\) and \(CDA = FDA => ACDF\) is parallelogram with equal sides => \(AC = X\) and \(BC = 18 - X\)
Now our question is the area of triangle CFD: the area of this triangle equals to the area of triangle ACD which equals to the difference between areas of triangle ABD and triangle ABC, lets find those.
triangle ABD: \(12 * 18 * 0.5 = 108\)
for triangle ABC: hypothenuse = \(x\), legs are \(18 - x\) and \(12\)
\((18-x)^2 + 144 = x^2\)
\(324 - 36*x + x^2 + 144 = x^2\)
\(X = 13\) which means that \(BC = 5\) and that means that area of triangle ABC = \(12*5*0.5 = 30\) which means that the resulting area is \(108 - 30 = 78\)

B

Bunuel Can we even expect such questions on the GMAT even if I consider it to be a 51 level geometry question? Apparently, what I have seen is that the GMAT tests more of logic than laborious method/calculations.

Can we have your view on the same. Thanks.

Just adding my thoughts here.

I would imagine that it is not totally out of the question. It all boils down to how quickly you can figure out the pythagorean relation for sides in triangle DEF (which requires logical deduction).

From the question we know, \(AF = DF\) and that angle DEF = 90, so for the triangle DEF, we can write:

\(=> AE^2 + (18-DF)^2 = DF^2\) --- this is the key relation that you have to find out; from this point onwards it should take about 30 sec to solve the rest and get to the answer.
\(=> 12^2 + (18-DF)^2 = DF^2\)
\(=> DF = \frac{12^2}{2*18} + \frac{18^2}{2*18}\)
(No laborious calculations here as DF^2 cancels out and you do not even need to find values of 12^2, 18^2 or 2*18. Just leave them as is and cancel out all the common factors and you will be left with 4 and 9)
\(=> DF = 4 + 9\)
\(=> DF = 13\)

Once you have DF, you can apply \(\frac{AB*DF}{2}\) to get the area of the triangle in question. (Another deduction required here is that height of the triangle is AB)
\(=> \frac{12*13}{2}\)

\(=> 78\)

Answer is B 78 is the answer

Good One Sahil...excellent approach..

BC+CD=18
and sum of their squares is CD square pythogoras theorm
BE=5
CD=13
similarly same approach for AF and FE
AF=13 and FE =5

area = 0.5x12x13=78

hey can u help me with my query?
I solved till getting x=5 but then i took area of small two triangles (not shaded) and tried subtracting it from area of rectangle, which come like 216-30-30 = 156
so where am I going wrong?

simply enough, A and D reflect through CF, but ABCD is rectangle => AFDC is rhombus.
Also, to find the value of area of shaded triangle, we need to know AF.
Use 2 numbers 12 & 18 => B

Bunuel
can you please explain how exactly the paper has been folded? has it been folded twice? The question does not look tough, but imagining how exactly the folding process took place is giving me nightmares. Please explain this. And if there are other questions like this out there then please also point them out. Thanks for your help.

I don;t know what can I add to what is already given there: The sheet is folded along the segment CF so that points A and D coincide after the paper is folded, as shown in Figure 2 (The shaded area represents a portion of the back side of the paper, not visible in Figure 1).

the key attack is to spot that DF = AF = CD => ACDF is a rhombus.

From a quick peek, can someone elaborate why CF divides the rectangle to supposedly two symmetrical trapezoids?
CF can also be more to the left or to the right.
Any clarification?

As we can see in figure 2 that DF=AF, AC=CD => angle DAF=ADF, CAD=CDA. And then we have BAC+CAD+DAF=FDE+ADF+CDA=90 ==> BAC=FDE =>BC=FE

We can let BC = x and thus CD = 18 - x. We see that triangle DBC (in figure 2) is a right triangle and thus we have:

(DB)^2 + (BC)^2 = (CD)^2

12^2 + x^2 = (18 - x)^2

144 + x^2 = 324 - 36x + x^2

36x = 180

x = 5

Likewise, if we let FE = y and thus AF = 18 - y. We see that triangle AEF (in figure 2) is a right triangle. Since side AE of triangle AEF is now same as DE, AE = 12 and we have:

(AE)^2 + (FE)^2 = (AF)^2

12^2 + y^2 = (18 - y)^2

This is equivalent to the equation above, so y = 5.

Now we can argue that:

Area of triangle DBC + Area of triangle AEF + 2(Area of triangle CAF) = Area of rectangle ABDE

(½)(12)(5) + (½)(12)(5) + 2(Area of triangle CAF) = 12(18)

30 + 30 + 2(Area of triangle CAF) = 216

2(Area of triangle CAF) = 156

Area of triangle CAF = 78

But the area of triangle CAF is the area of the shaded region, so the area of the shaded region is 78.

Answer: B

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