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# In the diagram above, figure ABCD is a square with an area of 4.5 in^2

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Math Expert
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Posts: 47084
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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54% (02:58) correct 46% (03:30) wrong based on 154 sessions

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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:

2015-06-02_1741.png [ 19.3 KiB | Viewed 3971 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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04 Jun 2015, 10:07
5
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=DT-DQ=1.5-1=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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2015-06-02_1741.png [ 24.11 KiB | Viewed 3253 times ]

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Manager
Joined: 12 Nov 2014
Posts: 62
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 45-45-90 triangle.
From here it is just application of Pythagoras theorem.
QC = sqrt {(1/√2)^2 + (2/√2)^2
= √10 / 2

Attachments

tri.jpg [ 16.93 KiB | Viewed 3379 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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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??

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 45-45-90 triangle.
From here it is just application of Pythagoras theorem.
QC = sqrt {(1/√2)^2 + (2/√2)^2
= √10 / 2

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Joined: 04 May 2014
Posts: 29
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.

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
1
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??

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 45-45-90 triangle.
From here it is just application of Pythagoras theorem.
QC = sqrt {(1/√2)^2 + (2/√2)^2
= √10 / 2

Manager
Joined: 12 Nov 2014
Posts: 62
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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Posts: 47084
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 2015-06-02_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:

Attachment:

2015-06-08_1619.png [ 24.81 KiB | Viewed 3438 times ]

Attachment:

2015-06-08_1619_001.png [ 24.45 KiB | Viewed 3446 times ]

Attachment:

2015-06-08_1622.png [ 225.92 KiB | Viewed 3469 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=DT-DQ=1.5-1=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
1
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:
2015-06-02_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:

Attachment:
2015-06-08_1619.png

Attachment:
2015-06-08_1619_001.png

Attachment:
2015-06-08_1622.png
Math Expert
Joined: 02 Sep 2009
Posts: 47084
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
1
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:
2015-06-02_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:

Attachment:
2015-06-08_1619.png

Attachment:
2015-06-08_1619_001.png

Attachment:
2015-06-08_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:
2015-06-02_1741.png

Hi Bunuel, chetan2u,

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
1
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:
2015-06-02_1741.png

Hi Bunuel, chetan2u,

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 180-75 =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   [#permalink] 28 Dec 2017, 02:17
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