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In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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02 Jun 2015, 06:45
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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02 Jun 2015, 10:01
Lets say DQ = x, QB = 2x Diagonal = 3x Side of the square = 3x/√2 Area = (3x/√2)^2 = 9x^2/2 = 4.5 9x^2 = 9; x =1 DQ = 1; QB = 2 ; Diagonal = 3 Side of square = 3/√2 See the attached figure. Since BD is diagonal, angle BDC is 45 degree. If you drop a perpendicular from Q to base CD you get a 454590 triangle. From here it is just application of Pythagoras theorem. QC = sqrt {(1/√2)^2 + (2/√2)^2 = √10 / 2 Answer A
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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04 Jun 2015, 09:21
I got the first half of when you mentioned although post this, I used normal Pythogoras QD^2 +QC^2 = CD^2 1 +QC^2=9/2 QC => \sqrt{7/2} what am I doing wrong here?? askhere wrote: Lets say DQ = x, QB = 2x Diagonal = 3x Side of the square = 3x/√2 Area = (3x/√2)^2 = 9x^2/2 = 4.5 9x^2 = 9; x =1 DQ = 1; QB = 2 ; Diagonal = 3 Side of square = 3/√2
See the attached figure. Since BD is diagonal, angle BDC is 45 degree. If you drop a perpendicular from Q to base CD you get a 454590 triangle. From here it is just application of Pythagoras theorem. QC = sqrt {(1/√2)^2 + (2/√2)^2 = √10 / 2
Answer A



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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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04 Jun 2015, 10:07
since the area is 4.5 or 45/10, each side of the square=\(\sqrt{\frac{45}{10}}\) = \(\frac{3}{\sqrt{2}}\) With sides \(\frac{3}{\sqrt{2}}\), diagonal will be 3.. join C with centre of diag at T.. CT will be 1.5...and QT=DTDQ=1.51=0.5.. now in triangle CTQ, CQ = \(\sqrt{(0.5)^2+(1.5)^2}\) = \(\sqrt{\frac{1}{4}+\frac{9}{4}}\) =\(\sqrt{10}/2\)... ans A
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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04 Jun 2015, 10:30
I seem to get lost here a bit. What I did first was to get the diagonal DB, since we know the area of 4.5 inch.
area of square = side^2 so the side would be sqrt (4.5) So Diagonal is sqrt( 4.5 )^2 + sqrt( 4.5 )^2 = sqrt ( 9 ) = 3
Since DQ ratio is 1:2, DQ has to be 1 This is the point I got lost on.
After some reading on another answer I thought the perpendicular line would help, but then i got stuck on how to simplify the square roots in a fraction. If someone could also help me out on some reading material regarding square roots and fractions and the reciprocals of them that would be great.
Because why is (From Askhere) QC = sqrt {(1/√2)^2 + (2/√2)^2 = √10 / 2 I mean when raising these to the power you would square root of 3 right?



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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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04 Jun 2015, 11:11
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QC is not perpendicular to DB.(see ratio DQ/QB is 1/2) so you can not apply Pythagoras here. Regards jimmy kelvind13 wrote: I got the first half of when you mentioned although post this, I used normal Pythogoras QD^2 +QC^2 = CD^2 1 +QC^2=9/2 QC => \sqrt{7/2} what am I doing wrong here?? askhere wrote: Lets say DQ = x, QB = 2x Diagonal = 3x Side of the square = 3x/√2 Area = (3x/√2)^2 = 9x^2/2 = 4.5 9x^2 = 9; x =1 DQ = 1; QB = 2 ; Diagonal = 3 Side of square = 3/√2
See the attached figure. Since BD is diagonal, angle BDC is 45 degree. If you drop a perpendicular from Q to base CD you get a 454590 triangle. From here it is just application of Pythagoras theorem. QC = sqrt {(1/√2)^2 + (2/√2)^2 = √10 / 2
Answer A



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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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04 Jun 2015, 20:39
[quote="kelvind13"]I got the first half of when you mentioned although post this, I used normal Pythogoras QD^2 +QC^2 = CD^2 1 +QC^2=9/2 QC => \sqrt{7/2} what am I doing wrong here?? Hi kelvind13, To apply Pythagoras theorem it has to be a right triangle. You have assumed Traingle DQC to be a right triangle which is wrong. This is where you went wrong. Hope you understood.
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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08 Jun 2015, 05:25
Bunuel wrote: In the diagram above, figure ABCD is a square with an area of 4.5 in^2. If the ratio of the length of DQ to the length of QB is 1 to 2, what is the length of QC, in inches? A. \(\frac{\sqrt{10}}{2}\) B. \(\frac{\sqrt{14}}{2}\) C. \(2\sqrt{2}\) D. \(2\sqrt{3}\) E. \(2\sqrt{5}\) Attachment: The attachment 20150602_1741.png is no longer available MANHATTAN GMAT OFFICIAL SOLUTION:If the area of the square is 4.5 and the side length is s, then \(s^2=\frac{9}{2}\) and \(s =\frac{3\sqrt{2}}{2}\) Triangle BCD is a 45–45–90 right triangle, so the ratio of its sides is \(s : s : s \sqrt{2}\). Thus, DB (the hypotenuse) is \(s \sqrt{2}=\frac{3\sqrt{2}}{2}*\sqrt{2}=3\). Furthermore, if DB is 3, and the ratio DQ : QB is 1 : 2, it must be true that DQ = 1 and QB = 2. To be able to use Pythagorean Theorem, we must create a right triangle with QC as a side by drawing a new line. Such a right triangle will be similar to triangle BCD. Here are two solutions, the only difference being how one draws the extra line: The correct answer is A.Attachment:
20150608_1619.png [ 24.81 KiB  Viewed 3240 times ]
Attachment:
20150608_1619_001.png [ 24.45 KiB  Viewed 3247 times ]
Attachment:
20150608_1622.png [ 225.92 KiB  Viewed 3266 times ]
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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02 Jul 2015, 04:17
Hey, can someone pls explain what is wrong with the below approach?
QD^2 +QC^2 = CD^2 1 +QC^2=9/2 QC => \sqrt{7/2}



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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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14 Mar 2017, 18:13
chetan2u wrote: since the area is 4.5 or 45/10, each side of the square=\(\sqrt{\frac{45}{10}}\) = \(\frac{3}{\sqrt{2}}\) With sides \(\frac{3}{\sqrt{2}}\), diagonal will be 3.. join C with centre of diag at T.. CT will be 1.5...and QT=DTDQ=1.51=0.5.. now in triangle CTQ, CQ = \(\sqrt{(0.5)^2+(1.5)^2}\) = \(\sqrt{\frac{1}{4}+\frac{9}{4}}\) =\(\sqrt{10}/2\)... ans A Hello Chetan, Can you please explain how did you concluded BT as 1.5?
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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27 Dec 2017, 13:07
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Dear Bunuel One thing that I didn't understood from the second Manhattam Official Solution is: why DQ = 2, as the ratio of both triangles is 1:3? Could you clarify that to me? Thanks! Bunuel wrote: Bunuel wrote: In the diagram above, figure ABCD is a square with an area of 4.5 in^2. If the ratio of the length of DQ to the length of QB is 1 to 2, what is the length of QC, in inches? A. \(\frac{\sqrt{10}}{2}\) B. \(\frac{\sqrt{14}}{2}\) C. \(2\sqrt{2}\) D. \(2\sqrt{3}\) E. \(2\sqrt{5}\) Attachment: 20150602_1741.png MANHATTAN GMAT OFFICIAL SOLUTION:If the area of the square is 4.5 and the side length is s, then \(s^2=\frac{9}{2}\) and \(s =\frac{3\sqrt{2}}{2}\) Triangle BCD is a 45–45–90 right triangle, so the ratio of its sides is \(s : s : s \sqrt{2}\). Thus, DB (the hypotenuse) is \(s \sqrt{2}=\frac{3\sqrt{2}}{2}*\sqrt{2}=3\). Furthermore, if DB is 3, and the ratio DQ : QB is 1 : 2, it must be true that DQ = 1 and QB = 2. To be able to use Pythagorean Theorem, we must create a right triangle with QC as a side by drawing a new line. Such a right triangle will be similar to triangle BCD. Here are two solutions, the only difference being how one draws the extra line: The correct answer is A.Attachment: 20150608_1619.png Attachment: 20150608_1619_001.png Attachment: 20150608_1622.png



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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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27 Dec 2017, 23:24
robertops wrote: Dear Bunuel One thing that I didn't understood from the second Manhattam Official Solution is: why DQ = 2, as the ratio of both triangles is 1:3? Could you clarify that to me? Thanks! Bunuel wrote: Bunuel wrote: In the diagram above, figure ABCD is a square with an area of 4.5 in^2. If the ratio of the length of DQ to the length of QB is 1 to 2, what is the length of QC, in inches? A. \(\frac{\sqrt{10}}{2}\) B. \(\frac{\sqrt{14}}{2}\) C. \(2\sqrt{2}\) D. \(2\sqrt{3}\) E. \(2\sqrt{5}\) Attachment: 20150602_1741.png MANHATTAN GMAT OFFICIAL SOLUTION:If the area of the square is 4.5 and the side length is s, then \(s^2=\frac{9}{2}\) and \(s =\frac{3\sqrt{2}}{2}\) Triangle BCD is a 45–45–90 right triangle, so the ratio of its sides is \(s : s : s \sqrt{2}\). Thus, DB (the hypotenuse) is \(s \sqrt{2}=\frac{3\sqrt{2}}{2}*\sqrt{2}=3\). Furthermore, if DB is 3, and the ratio DQ : QB is 1 : 2, it must be true that DQ = 1 and QB = 2. To be able to use Pythagorean Theorem, we must create a right triangle with QC as a side by drawing a new line. Such a right triangle will be similar to triangle BCD. Here are two solutions, the only difference being how one draws the extra line: The correct answer is A.Attachment: 20150608_1619.png Attachment: 20150608_1619_001.png Attachment: 20150608_1622.png It's a typo there. DQ should be 1.
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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28 Dec 2017, 01:24
Bunuel wrote: In the diagram above, figure ABCD is a square with an area of 4.5 in^2. If the ratio of the length of DQ to the length of QB is 1 to 2, what is the length of QC, in inches? A. \(\frac{\sqrt{10}}{2}\) B. \(\frac{\sqrt{14}}{2}\) C. \(2\sqrt{2}\) D. \(2\sqrt{3}\) E. \(2\sqrt{5}\) Attachment: 20150602_1741.png Hi Bunuel, chetan2u, Can you please help me to clarify my doubt. From the given info on the question DB = 3, DQ =1 & QB = 2. Let P be mid point of QB so, DQ = QP = PB = 1. As each side are equal, DQ, QP & PB will subtend an equal angle at C so Angle QCP = Angle QCD = Angle PCB = 30. Now, as per property of Triangle for Triangle QCD \(\frac{QD}{Sin30} = \frac{QC}{Sin45} = \frac{CD}{Sin75}\). By using this logic I am getting a different answer for QC. Can you please guide as to where I am going wrong.
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2 [#permalink]
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28 Dec 2017, 02:17
rahul16singh28 wrote: Bunuel wrote: In the diagram above, figure ABCD is a square with an area of 4.5 in^2. If the ratio of the length of DQ to the length of QB is 1 to 2, what is the length of QC, in inches? A. \(\frac{\sqrt{10}}{2}\) B. \(\frac{\sqrt{14}}{2}\) C. \(2\sqrt{2}\) D. \(2\sqrt{3}\) E. \(2\sqrt{5}\) Attachment: 20150602_1741.png Hi Bunuel, chetan2u, Can you please help me to clarify my doubt. From the given info on the question DB = 3, DQ =1 & QB = 2. Let P be mid point of QB so, DQ = QP = PB = 1. As each side are equal, DQ, QP & PB will subtend an equal angle at C so Angle QCP = Angle QCD = Angle PCB = 30. Now, as per property of Triangle for Triangle QCD \(\frac{QD}{Sin30} = \frac{QC}{Sin45} = \frac{CD}{Sin75}\). By using this logic I am getting a different answer for QC. Can you please guide as to where I am going wrong. You have gone wrong in taking all three angles as 30.. If you have got angle QCD as 75 so QCB as 18075 =105... Now if I take the centre triangle QCP, it should be isosceles as the side triangles QCD and PCB are similar.. So the two angles become 105*2=210>180.. not possible..
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Re: In the diagram above, figure ABCD is a square with an area of 4.5 in^2
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