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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 = FD. Find the area of the shaded figure. A. 1/3 B. 13/16 C. 5/9 D. 1/2 Attachment:
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Originally posted by RaviChandra on 14 Sep 2010, 22:58.
Last edited by Bunuel on 06 May 2020, 07:59, edited 3 times in total.
Renamed the topic and edited the question.



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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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14 Sep 2010, 22:59
In triangle ABC, D and E are points on AC and AB such DE  BC and length of DE is onethird 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 16027 times ]



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ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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14 Sep 2010, 23:54
RaviChandra wrote: 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 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. Attachment:
7.PNG [ 5.36 KiB  Viewed 15985 times ]
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.
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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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15 Sep 2010, 00:39
Wht is answer of the 2nd question ???
24 ?



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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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15 Sep 2010, 01:03
RaviChandra wrote: In triangle ABC, D and E are points on AC and AB such DE  BC and length of DE is onethird 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 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 [same height, 1/3rd base] 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)=21624=192 So x+5y=96 Subtracting the 2 equations 4y=48 y=12 Hence, Area(DEF)=12
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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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15 Sep 2010, 04:02
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



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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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28 Sep 2015, 05:38
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 midpoint 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.
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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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06 May 2020, 02:18
MentorTutoringCan you pls explain how we ascertain height of shaded triangle relative to triangle ABC in this question



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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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06 May 2020, 05:13
GDT wrote: MentorTutoringCan you pls explain how we ascertain height of shaded triangle relative to triangle ABC in this question 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 threestep 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 testtakers 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



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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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06 May 2020, 05:33
MentorTutoring wrote: GDT wrote: MentorTutoringCan you pls explain how we ascertain height of shaded triangle relative to triangle ABC in this question 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 threestep 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 testtakers 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 Hi MentorTutoringI 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!



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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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06 May 2020, 06:36
GDT wrote: Hi MentorTutoringI 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! 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



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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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06 May 2020, 07:29
MentorTutoring wrote: GDT wrote: Hi MentorTutoringI 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! 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 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?



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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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06 May 2020, 09:26
GDT wrote: 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?
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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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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06 May 2020, 23:01
MentorTutoring wrote: GDT wrote: 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?
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 Thank you! Now, it's clear



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Re: ABC is a triangle with area 1. AF = AB/3, BE = BC/3 and ED =
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07 May 2020, 07:14
GDT wrote: MentorTutoring wrote: GDT wrote: 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?
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 Thank you! Now, it's clear That is music to my ears, GDT. Glad I could borrow a key "prop," as it were, from Bunuel. To reconstruct that same diagram in the way I know would take a long time. I will scroll back up to give that post a kudos right after this. Good luck with your studies.  Andrew




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