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# Segment BDF bisects both square ABCD and square DEFG and has a length

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Posts: 58464
Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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23 Aug 2018, 03:56
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Difficulty:

45% (medium)

Question Stats:

67% (02:21) correct 33% (02:43) wrong based on 63 sessions

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Segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units. If the ratio of the length of a side of ABCD to a side of DEFG is 3 : 4, what is the sum of the areas of the squares?

A. 196
B. 100
C. 64
D. 52
E. 36

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Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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Updated on: 23 Aug 2018, 06:25
Bunuel wrote:
Segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units. If the ratio of the length of a side of ABCD to a side of DEFG is 3 : 4, what is the sum of the areas of the squares?

A. 196
B. 100
C. 64
D. 52
E. 36

Let x and y are the length of the sides of the squares ABCD & DEFG respectively.

Given x:y=3k:4k
Given, segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units
Or, BD+DF=$$14\sqrt{2}$$
Or, $$\sqrt{2}(x+y)$$=$$14\sqrt{2}$$
Or, x+y=14
or, 3k+4k=14
Or, k=2

So, x=3k=6, y=4k=8

So sum of area of both the squares=$$6^2+8^2=100$$

Ans. (B)
Edit: Fixed the mathematical formula

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Originally posted by PKN on 23 Aug 2018, 04:59.
Last edited by PKN on 23 Aug 2018, 06:25, edited 1 time in total.
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Posts: 252
Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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23 Aug 2018, 06:17
PKN wrote:
Bunuel wrote:
Segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units. If the ratio of the length of a side of ABCD to a side of DEFG is 3 : 4, what is the sum of the areas of the squares?

A. 196
B. 100
C. 64
D. 52
E. 36

Let x and y are the length of the sides of the squares ABCD & DEFG respectively.

Given x:y=3k:4k
Given, segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units
Or, BD+DF=$$14\sqrt{2}$$
Or, \sqrt{x}+[fraction]y[/fraction]=$$14\sqrt{2}$$
Or, x+y=14
or, 3k+4k=14
Or, k=2

So, x=3k=6, y=4k=8

So sum of area of both the squares=$$6^2+8^2=100$$

Ans. (B)
Hi PKN,
Can you please show the diagram of the question stem. I am not able to comprehend the question stem.

Sumit

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Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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23 Aug 2018, 06:49
sumit411 wrote:
PKN wrote:
Bunuel wrote:
Segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units. If the ratio of the length of a side of ABCD to a side of DEFG is 3 : 4, what is the sum of the areas of the squares?

A. 196
B. 100
C. 64
D. 52
E. 36

Let x and y are the length of the sides of the squares ABCD & DEFG respectively.

Given x:y=3k:4k
Given, segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units
Or, BD+DF=$$14\sqrt{2}$$
Or, \sqrt{x}+[fraction]y[/fraction]=$$14\sqrt{2}$$
Or, x+y=14
or, 3k+4k=14
Or, k=2

So, x=3k=6, y=4k=8

So sum of area of both the squares=$$6^2+8^2=100$$

Ans. (B)
Hi PKN,
Can you please show the diagram of the question stem. I am not able to comprehend the question stem.

Sumit

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

Hi sumit411,

I have enclosed the required diagram.
Thank you.
Note:- Since x and y are of different ratio, hence squares are not identical though it seems identical in figure.
Attachments

Double square.JPG [ 18.6 KiB | Viewed 1189 times ]

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Regards,

PKN

Rise above the storm, you will find the sunshine
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Posts: 252
Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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24 Aug 2018, 01:52
PKN wrote:
sumit411 wrote:
PKN wrote:
[quote="Bunuel"]Segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units. If the ratio of the length of a side of ABCD to a side of DEFG is 3 : 4, what is the sum of the areas of the squares?

A. 196
B. 100
C. 64
D. 52
E. 36

Let x and y are the length of the sides of the squares ABCD & DEFG respectively.

Given x:y=3k:4k
Given, segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units
Or, BD+DF=$$14\sqrt{2}$$
Or, \sqrt{x}+[fraction]y[/fraction]=$$14\sqrt{2}$$
Or, x+y=14
or, 3k+4k=14
Or, k=2

So, x=3k=6, y=4k=8

So sum of area of both the squares=$$6^2+8^2=100$$

Ans. (B)
Hi PKN,
Can you please show the diagram of the question stem. I am not able to comprehend the question stem.

Sumit

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

Hi sumit411,

I have enclosed the required diagram.
Thank you.
Note:- Since x and y are of different ratio, hence squares are not identical though it seems identical in figure.[/quote]Thanks for your effort
Surely helps.

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
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Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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26 Aug 2018, 19:23
Bunuel wrote:
Segment BDF bisects both square ABCD and square DEFG and has a length of $$14\sqrt{2}$$ units. If the ratio of the length of a side of ABCD to a side of DEFG is 3 : 4, what is the sum of the areas of the squares?

A. 196
B. 100
C. 64
D. 52
E. 36

We see that 14√2 is the sum of the lengths of the diagonal of square ABCD and the diagonal of square DEFG. Since the ratio of the sides of the two squares is 3:4, the ratio of the diagonals of the two squares is also 3:4. So we can let the diagonal of square ABCD = 3x and the diagonal of square DEFG = 4x and we have:

3x + 4x = 14√2

7x = 14√2

x = 2√2

Therefore, the diagonal of square ABCD is 3(2√2) = 6√2 and hence its side = 6 and its area = 6^2 = 36. Similarly, the diagonal of square DEFG is 4(2√2) = 8√2 and hence its side = 8 and its area = 6^2 = 64. So the sum of the areas of the two squares is 36 + 64 = 100.

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Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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27 Aug 2018, 02:23
1
Using the unkown multiplier approach:

according to the formula for 45-45-90 triangles, the side opposite to the 90° angle = $$x\sqrt{2}$$ = 14$$\sqrt{2}$$
thus the length of the 2 triangles combined = 14

ratio of the sides of the 2 sqares: $$3x/4x-> total = 7x$$
7x = 14
x=2 = unkown multiplier

sides lengths are:
$$2*3 and 2*4$$

Areas of both squares $$= 6^2+8^2 = 100$$

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Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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27 Aug 2018, 10:42
How do we know that the two squares only intersect at one point (D)?

Given the conditions, is it possible that the two squares share point D and overlap at two sides? Thus the given diagonal 14(root2) would simply be the diagonal of the larger square?

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Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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27 Aug 2018, 11:28
hastee wrote:
How do we know that the two squares only intersect at one point (D)?

Given the conditions, is it possible that the two squares share point D and overlap at two sides? Thus the given diagonal 14(root2) would simply be the diagonal of the larger square?

Posted from my mobile device

Hi hastee,

Could you please enclose the diagram you have cited above?
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PKN

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Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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27 Aug 2018, 12:01
I have too few posts to link an image.

My question: what verbiage in the question tells us that squares only intersect at point D. My concern is that 3 vertices of the smaller square may lie on the larger square, with the 4th point of the smaller square contained within the larger square.

The larger DEFG would contain, or intersect all points of square ABCD.
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Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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27 Aug 2018, 12:02
Now, I have enough posts

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Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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27 Aug 2018, 12:34
1
hastee wrote:
Now, I have enough posts

Given in the question:-
Segment BDF has a length of $$14\sqrt{2}$$ unit.
Here two cases can be concluded from the nomenclature of the segment BDF. viz.,
1) B - D - F
2) F - D - B

In succinct, D lies in between B and F or F and B.

According to your figure neither of the above terminology meets. Rather 'D' is at one corner making D_B_F or F_B_D. where B lies in between F and D.
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Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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29 Aug 2018, 12:25
I made this diagram because someone mentioned they diagram already posted by someone else had the squares look the same size.

Though, you don't really have to go to all that of a hassle to solve this question.

Posted from my mobile device
Attachments

File comment: Diagram of squares with bisecting segment

IMG_20180830_002115.jpg [ 4.36 MiB | Viewed 828 times ]

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Re: Segment BDF bisects both square ABCD and square DEFG and has a length  [#permalink]

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Re: Segment BDF bisects both square ABCD and square DEFG and has a length   [#permalink] 20 Oct 2019, 18:02
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