In the figure above, DE parallels BA, and DF parallels CA. What is the

Question

df.png In the figure above, DE parallels BA, and DF parallels CA. What is the area of the triangle ABC? (1) Area of quadrilateral AEDF is 40. (2) Area of triangle BDF is 50, and area of triangle DEC is 8.

vitaliyGMAT · Answer

Hi

(1) Even identifying different possible values of FD, DE we won't get EC or BD or at least relationship of side for similar triangles DEC and BAC. Insufficient.

(2) Triangles BFD, DEC and BAC are similar.

\frac{Area_{BDF}}{Area_{DEC}} = \frac{50}{8} = \frac{25}{4} = k^2

k = \frac{5}{2}

BA = \frac{5}{2}AF + AF = \frac{7}{2}AF

BC = \frac{5}{2}DC + DC = \frac{7}{2}DC

...

\frac{BC}{DC} = \frac{7}{2} = k^'

\frac{Area_{BAC}}{Area_{DEC}} = (k^')^2 = \frac{49}{4}

Area_{BAC} = \frac{49}{4}*8 = 98

Sufficient.

Answer B

rohit8865 · Answer

(1) clearly insuff

(2) BDF & DEC are similar triangles 
thus Area of BDC/Area DEC= BF^2/DE^2
BF/DE= 5 √2/2 √2= 5/2
BF = 5DE/2
Also DE= AF (4 lines given parallel)

AB = BF+AF==BF+DE = 5DE/2 +DE= 7DE/2

Now DEC & ABC are similar triangles so

Area ABC / Area DEC = AB^2 /DE^2= (7DE/2)^2/ DE^2= 49/4
Area ABC / 8 = 49/4 (as Area DEC=8)
Area ABC = 49*2 = 98 units^2
suff

Ans B

ccooley · Answer

My thoughts:

First, it  definitely  isn't C. You should realize that within the first 30 seconds. If it seems obvious that you can solve with both statements together, C isn't the right answer (it's a "C trap"). Likewise, since it's possible to solve with both together, the answer can't be E. We're left with A or B, or (less likely) D. That means we'll be doing something sort of unusual on this problem: we're going to focus mostly on showing that the statements are  sufficient , instead of proving that they're  insufficient , like you would on most DS problems. We're doing this because we already know, via logic, that there's at least one statement that's sufficient by itself. We just have to figure out which one it is (or whether it's both).

(2) seems to give me more information, so I'll start there.

(2) Notice that the two triangles are similar. (You can prove this, or you can memorize this diagram as an example of similar triangles.)

That means that the height and base of the larger triangle are proportional to the height and base of the smaller triangle.

https://i.imgur.com/cvigInT.png

('x' is the factor by which they're proportional. I don't really care what it is! What matters is that we  could  calculate that proportion, since we have the areas of both triangles.)

With everything labeled like this, we can find the area of the biggest triangle: 0.5 * (x*base + base) * (x*height + height). Simplify.

0.5 * (x + 1) * base * (x + 1) * height

(x + 1)^2 * 0.5 * base * height

We already know 0.5 * base * height; that's the area of the smaller triangle, which is 8. We could also calculate (x+1)^2, since we know x. So, we could find the area of the biggest triangle.  Sufficient.

(1) This one might or might not also be sufficient. Let's figure it out. On most geometry DS problems, if I have an uncomplicated statement, I like to start with 'rubber band geometry' - draw out two different cases and see if they come out the same, or different.

Here's one possibility:

https://i.imgur.com/TAnaoUr.png

Here's another:

https://i.imgur.com/Q5vvjct.png

Insufficient.

Notice that I've made one strategic 'mistake' here: I acted as if I knew that all of the triangles were right triangles. Bonus points for anybody who can explain why that's actually okay, even though we don't usually want to make assumptions on DS problems!

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In the figure above, DE parallels BA, and DF parallels CA. What is the

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