ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =

Question

https://gmatclub.com/forum/download/file.php?id=12722 ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED = FD. Find the area of the shaded figure. A. 1/3 B. 13/16 C. 5/9 D. 1/2 1.JPG

RaviChandra · Answer

In triangle ABC, D and E are points on AC and AB such DE || BC and length of DE is one-third of BC. If the area of triangle ABC is 216 square units, find the area of the shaded triangle.
a. 12
b. 18
c. 24
d. 16

Attachment:

2.PNG [ 2.36 KiB | Viewed 27262 times ]

Bunuel · Answer

Rotate the triangle so that CA to be the base. Now, triangle CDA has the same base as CBA and half of its height ("read" the diagram for explanation), which means that area of CDA will be half of the area of CBA, so 1/2. 7.PNG Answer: D. RaviChandra these are not GMAT questions, so I wouldn't worry about them too much and definitely wouldn't spend much time on them.

Dohalad · Answer

Wht is answer of the 2nd question ???

24 ?

shrouded1 · Answer

Note that AED is similar to ABC with each side as 1/3rd of the original
=> Area(AED) = 1/9 * Area(ABC) =24

Area(ABD) = Area(ACE) = 1/3 * Area(ABC) = 72

Area(ABD) = Area(ACE)
=> Area(ABD) - Area(AEDF) = Area(ACE) - Area(AEDF)
=> Area(BEF) = Area(CDF) = x (say)

Finally note that DEF is similar to BFC with one side in the ratio of 1:3
Thus area(DEF) = 1/9 * area(BFC) = y (say)

Area(EDC) = Area(EDF)+Area(DFC) = x + y 
ALSO
Area(EDC) = Area(AEC) - Area(AED) =72 - 24 = 48
So x+y = 48

Area(BEDC) = Area(BEF) + Area(CDF) + Area(EDF) + Area(BFC) = x+x+y+9y 
ALSO
Area(BEDC) = Area(ABC)-Area(AED)=216-24=192
So x+5y=96

Subtracting the 2 equations
4y=48
y=12
 
 Hence, Area(DEF)=12

Dohalad · Answer

Wow... thks a ton.

I had guessed it to be 24 - Thinking T. EFD should be of area 1/9th of main traingle.

If I knew T. AED = 1/9 T. ABC, 12 would have been my guess, as it (T. EFD) looks half the size of T. AED

rohitmanglik · Answer

Whoa!!! A tough one

A small trick that helped me to solve it in 1:37

Imagine that the triangle ABC is an isosceles triangle with AB = BC. (Reason: All the mentioned conditions are satisfied so area ratio should also hold for Isosceles triangle)

Redraw it in your notepad, you will find symmetry. D is just in middle of AF and DE. You will figure out that D is a mid-point so its height should be half. Now if height of triangle is reduced to half, area of triangle will also reduce to half of its original value (Area = 0.5* base * height).

Hence D.

GDT · Answer

MentorTutoring

Can you pls explain how we ascertain height of shaded triangle relative to triangle ABC in this question

AndrewN · Answer

Thank you,  GDT , for tagging me. This is a visual puzzle more than anything, and to be honest, I cannot better demonstrate how to approach this question than   Bunuel   has done  above . If you rotate the figure and mark all the congruencies, you will see how the answer fits without knowing exact values. I will only add that a useful three-step approach to any Quant question is to ask yourself the following:

1) What is the question asking?

2) What information do I have to work with?

3) What is the core concept being tested?

In this instance, I think many test-takers would overlook the obvious in 3) above. The question is asking about  area , and the shape is a triangle, so the equation

A=\frac{1}{2}*b*h

will come into play. To make any sort of meaningful comparison, we need to find a way to hold one of our two variables constant while we manipulate the other. In terms of a guessing strategy, only (C) or (D) make any sense, given the setup. It is really down to whether you can find a way to justify half of the area of the overall triangle or slightly more than half.

- Andrew

GDT · Answer

Hi  MentorTutoring 
I read the explanation given by Bunuel but could not understand how we get to know that height of shaded triangle is half of that of triangle ABC

Could you pls explain how we get to infer that from the figure

Thanks in advance!

AndrewN · Answer

If you look at the left and right sides of the largest figure in the diagram, you can see how each third is congruent on that side. Since the problem tells us that ED = FD, because that diagonal cuts through the central third, we can deduce that the "height" of triangle ADC comprises  exactly  half of the middle third plus all of the bottom third, or one and a half out of three thirds of the largest triangle. How can you tell those thirds are congruent? It has to do with parallel lines (or segments) and, again, the information provided in the problem. If BE is 1/3 of BC and you drew a segment from point E that was parallel to AC, you would have 1/3 of side AB; likewise, if you drew a segment from point F that was parallel to AC, you have have 1/3 of side BC, since AF is 1/3 of AB. That leaves each side (that we are not calling the base) of the largest figure carved up into equal thirds. Does it make more sense now?

- Andrew

GDT · Answer

Is there any theorem that when a diagonal of a trapezium is getting bisected then at that point the ht. is divided into exactly 2 halves?

AndrewN · Answer

I am in the middle of a lesson, so I must be brief. I think you are looking for a definition more than a theorem:  midsegment .  Here  is a link you might find useful on the topic. I hope it helps.

- Andrew

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ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =

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ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =

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