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The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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09 Aug 2015, 08:33
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The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks
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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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09 Aug 2015, 12:11
reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks Hello retoThis task is built on isosceles triangle and its ratio \(1:1:\sqrt{2}\) \(EB=BF=3\) so this is isosceles triangle and we can infer that \(EF = 3\sqrt{2}\) So \(EG = GF = GB = 3\sqrt{2}/2\) Triangle DAB is isosceles so its side \(DB = 6\sqrt{2}\) \(DG = DB  GB\) > \(6\sqrt{2}  3\sqrt{2}/2 = 4.5\sqrt{2}\) Triangle DGH is isosceles so its side \(HG = 4.5\)  In your solution you are correctly find length of EG but we should subtract from AB not EG but IC which much less than EG.
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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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09 Aug 2015, 08:55
reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks You are assuming in your calculations that EG = GF .



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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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09 Aug 2015, 11:10
Engr2012 wrote: reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks You are assuming in your calculations that EG = GF . In triangle GFM with GM perpendicular to BF. The pain of nonnative. I confused intersect with bisect. Mh, what should I say, pretty important Thanks, lessons learned.
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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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09 Aug 2015, 11:15
reto wrote: Engr2012 wrote: reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks You are assuming in your calculations that EG = GF . The pain of nonnative. I confused intersect with bisect. Mh, what should I say, pretty important Thanks, lessons learned. I am not a big fan of ball parking in Quant as is done by Economist GMAT . I will ballpark/simplify my calculations but only based on some common sense rules. GNAT tutor's ballpaking methods are a bit more 'forced', IMO.



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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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09 Aug 2015, 12:05
Engr2012 wrote: reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks You are assuming in your calculations that EG = GF . Hello Engr2012And why this assuming is wrong?
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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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09 Aug 2015, 13:00
Harley1980 wrote: Engr2012 wrote: reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks You are assuming in your calculations that EG = GF . Hello Engr2012And why this assuming is wrong? Harley1980, my bad. Yes, it is infact correct that EG=GF.



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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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10 Aug 2015, 05:54
reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks Given: DC = 6, we need to find DI => DI = DC  IC We know that EB = BF = 3(Isosceles, 45:45:90), hence EF =3\(\sqrt{2}\). Extend HG to meet BC at O. Since angle BGF = 90, angle FGO = 45. Also GOF = 90, Hence angle OFG = 45. Traingle OGF is also a (45:45:90 triangle) Since GF = 3\(\sqrt{2}\) /2 => 1.5 \(\sqrt{2}\) => GO = OF = 1.5 We know that GO = IC = 1.5 Hence DI = 6 1.5 = 4.5 Option D



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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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18 Aug 2015, 11:28
Harley1980 wrote: reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks Hello retoThis task is built on isosceles triangle and its ratio \(1:1:\sqrt{2}\) \(EB=BF=3\) so this is isosceles triangle and we can infer that \(EF = 3\sqrt{2}\) So \(EG = GF = GB = 3\sqrt{2}/2\) Triangle DAB is isosceles so its side \(DB = 6\sqrt{2}\) \(DG = DB  GB\) > \(6\sqrt{2}  3\sqrt{2}/2 = 4.5\sqrt{2}\) Triangle DGH is isosceles so its side \(HG = 4.5\)  In your solution you are correctly find length of EG but we should subtract from AB not EG but IC which much less than EG. Hello! I am now reviewing this question. Why do you infer that EG = GF = GB ... I mean it is clear that EG = GF but why should it be also equal to the height of the triangle GB? Thank you Sergii!
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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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18 Aug 2015, 11:51
reto wrote: Harley1980 wrote: reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to know the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks Hello retoThis task is built on isosceles triangle and its ratio \(1:1:\sqrt{2}\) \(EB=BF=3\) so this is isosceles triangle and we can infer that \(EF = 3\sqrt{2}\) So \(EG = GF = GB = 3\sqrt{2}/2\) Triangle DAB is isosceles so its side \(DB = 6\sqrt{2}\) \(DG = DB  GB\) > \(6\sqrt{2}  3\sqrt{2}/2 = 4.5\sqrt{2}\) Triangle DGH is isosceles so its side \(HG = 4.5\)  In your solution you are correctly find length of EG but we should subtract from AB not EG but IC which much less than EG. Hello! I am now reviewing this question. Why do you infer that EG = GF = GB ... I mean it is clear that EG = GF but why should it be also equal to the height of the triangle GB? Thank you Sergii! In the text in red above, you mean why in triangle EBF, the height GB = EG = GF ? If this is the case then look below, You know in triangle EBF, we have 454590 triangle and angle EFB = 45. Draw the height GB such that angle BGF = 90. Thus in triangle BGF, angle BGF = 180angle GFB  angle BGF = 1809045 = 45 > BG = GF



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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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18 Aug 2015, 19:12
Engr2012 wrote: reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks You are assuming in your calculations that EG = GF . Assumption EG = GF looks valid because HGID is square and thus HD must be equal to DI, and hence AH and CI must be equal. so EG = GF must be true. Please correct me if i am wrong
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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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19 Aug 2015, 09:43
Engr2012 wrote: In the text in red above, you mean why in triangle EBF, the height GB = EG = GF ?
If this is the case then look below,
You know in triangle EBF, we have 454590 triangle and angle EFB = 45. Draw the height GB such that angle BGF = 90.
Thus in triangle BGF, angle BGF = 180angle GFB  angle BGF = 1809045 = 45 > BG = GF Hey! Thanks I got it now. But for me, impossible to see in 2min... now. Maybe after a lot more of practice. Better to rely on ballparking ...
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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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19 Aug 2015, 10:13
Everyone is making this problem way harder than it actually is. This problem does not require using sqrt(2) or any ballpark "guessing". If you split up the top triangle into a square and two triangles it should be easy to see the square has a side of 3/2 as well as the neighbor triangles. DI simply has length 63/2.



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The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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19 Aug 2015, 11:20
PolarCatastrophe wrote: Everyone is making this problem way harder than it actually is. This problem does not require using sqrt(2) or any ballpark "guessing". If you split up the top triangle into a square and two triangles it should be easy to see the square has a side of 3/2 as well as the neighbor triangles. DI simply has length 63/2. Yes, I missed the fact that HGID is a square as well.



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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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19 Aug 2015, 11:23
dav90 wrote: Engr2012 wrote: reto wrote: Attachment: T7914.png The side of square ABCD is 6. The midpoints of AB and BC are E and F, respectively. If square HGID intersects EF at G, what is the length of DI? A. 3 B. 4 C. 4.5 D. 5 E. 5.5 Economist says ballpark, but I would like to knoNw the calculations for this. What is wrong with my approach: square HGID intersects EF at G. Triangle EBF is a 904545 triangle and therefore the hypothenuse is: 3*sqrt(2). Half of 3*sqrt(2) would be a little more than 2. No lets subtract a little more than 2 from 6, which would equal a little less than 4. It does not make sense, because none of the answers are in this region. Thanks You are assuming in your calculations that EG = GF . Assumption EG = GF looks valid because HGID is square and thus HD must be equal to DI, and hence AH and CI must be equal. so EG = GF must be true. Please correct me if i am wrong Yes, you are correct.



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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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20 Aug 2015, 03:47
PolarCatastrophe wrote: Everyone is making this problem way harder than it actually is. This problem does not require using sqrt(2) or any ballpark "guessing". If you split up the top triangle into a square and two triangles it should be easy to see the square has a side of 3/2 as well as the neighbor triangles. DI simply has length 63/2. Wow, respect. Yes that seems like a useful approach. So expanding triangle EBF into a Square with side length of 6/2. Following this, half of the newly built square will be 3/2. KUDOS for that
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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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20 Aug 2015, 08:58
Even though if we make the traiangle to form a small square, where did 3/2 logic came from. Are you assuming "G" as the mid point of "EF". Pls explain.



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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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20 Aug 2015, 11:01
Shreeya24 wrote: Even though if we make the traiangle to form a small square, where did 3/2 logic came from. Are you assuming "G" as the mid point of "EF". Pls explain. Hello Shreeya24We are not assuming: G is the mid point of EF because DHGI is a square and E and F are midpoints of AB and BC Try to move point G and you will see that if G is not a midpoint of EF then DHGI is not a square. It is proved here: thesideofsquareabcdis6themidpointsofabandbcareeandf203250.html#p1558985But you can see it visually without any proves.
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Re: The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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21 Aug 2015, 09:32
Understood Thanks



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The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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21 May 2017, 13:51
Here's a faster explanation with almost no calculation involved, or what I thought:
GI splits EB in halves because of nature of the square which is symmetrical , therefor HG = 6  1/2 EB = 4.5.
Also, G must be the mid point of EF because if it's not, then you cannot fold the 2 squares into a bigger triangle with the biggest side being the diagonal.




The side of square ABCD is 6. The midpoints of AB and BC are E and F,
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