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

Question

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

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

PKN · Answer

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)
 Fixed the mathematical formula

sumit411 · Answer

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

Thanks in Advance.
Sumit

Sent from my ONEPLUS A3003 using  GMAT Club Forum mobile app

PKN · Answer

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.

sumit411 · Answer

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. Thanks for your effort

Surely helps. Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app

ScottTargetTestPrep · Answer

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.

Answer: B

laulau · Answer

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

Hit Kudos if you liked this answer

hastee · Answer

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

PKN · Answer

Hi  hastee ,

Could you please enclose the diagram you have cited above?

hastee · Answer

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.

hastee · Answer

Now, I have enough posts

https://i.imgur.com/yPCVEJL.jpg

PKN · Answer

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.

MsInvBanker · Answer

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

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

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

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