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# In the diagram above, ABCD is a square with side M, and EFGH is a squa

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In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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09 Apr 2015, 07:29
1
9
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Difficulty:

55% (hard)

Question Stats:

64% (02:36) correct 36% (02:40) wrong based on 184 sessions

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In the diagram above, ABCD is a square with side M, and EFGH is a square with side K. What is the value of (M + K)?

(1) (M – K) = 9

(2) The area of triangle AEH is 36

Kudos for a correct solution.

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gdrtq_img5.png [ 8.81 KiB | Viewed 4931 times ]

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In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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09 Apr 2015, 23:03
1
1
Bunuel wrote:

In the diagram above, ABCD is a square with side M, and EFGH is a square with side K. What is the value of (M + K)?

(1) (M – K) = 9

(2) The area of triangle AEH is 36

Kudos for a correct solution.

Attachment:
The attachment gdrtq_img5.png is no longer available

(1) (M – K) = 9
Area of one triangle is $$\frac{M^2 - K^2}{4}$$= $$\frac{(M+K)(M-K)}{4}$$ = $$\frac{9*(M+K)}{4}$$
we cannot inference much on $$\frac{9*(M+K)}{4}$$
Insufficient.

(2) The area of triangle AEH is 36
if AH=x
then (M-x)x = 72
again we cannot inference much .
insufficient.

together :

as we know area = $$\frac{9*(M+K)}{4}$$ = 36 . we can find (m+k)
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Re: In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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11 Apr 2015, 07:26
1
To find M+K

Area of shaded region =$$\frac{M^2 - K^2}{4}$$

We need to find M+K

1) M-K = 9 . Insufficient

2) $$\frac{M^2 - K^2}{4}$$ = 36 Not sufficent

From 1 and 2 we can get M+K. sufficient

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Re: In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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11 Apr 2015, 11:51
Hi Lucky2783

Could you please explain why the 4 triangles are congruent?
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Re: In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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11 Apr 2015, 16:10
1
Hi Naina1,

The reason why the 4 triangles are congruent involves a few geometry rules:

1) The angles on a line sum to 180 degrees
2) The angles in a triangle sum to 180 degrees
3) The corners of squares are 90 degree angles.

To prove the point, I'm going to have you do a little work....

1) Start with the purple triangle in the drawing and select two values (you get to choose) for the two non-90 degree angles. Remember that you're dealing with a triangle, so the sum of the angles has to be 180 degrees.
2) Next, working clockwise or counter-clockwise, figure out the next angle (remember, the corners of squares are 90 degrees and lines must sum to 180 degrees).

As you keep working, you'll find that each of the other triangles has the exact same angles (and side lengths) as the purple one.

GMAT assassins aren't born, they're made,
Rich
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Re: In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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11 Apr 2015, 20:36
Naina1 wrote:
Hi Lucky2783

Could you please explain why the 4 triangles are congruent?

one right angle.
one side of triangle i.e. Hypotnuse
and the highlighted angle are equal
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Re: In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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13 Apr 2015, 05:33
Bunuel wrote:

In the diagram above, ABCD is a square with side M, and EFGH is a square with side K. What is the value of (M + K)?

(1) (M – K) = 9

(2) The area of triangle AEH is 36

Kudos for a correct solution.

Attachment:
gdrtq_img5.png

VERITAS PREP OFFICIAL SOLUTION:

In the diagram, we literally have two squares, M squared and K squared. The four right triangle are what we get if we subtract one square from the other: they are literally the difference of two squares, so we can use the Difference of Two Squares formula:

M^2 - K^2 = (M + K)(M - K)

We could find (M + K) if we knew (M – K) and the difference of the squares.

Statement #1: this gives us (M – K), but we don’t know the difference of the two squares. This statement, alone and by itself, is not sufficient.

Statement #2: this gives us the area of one triangle, and if we multiply by 4, we have the difference of the two squares. But, now we don’t know (M – K), so we can solve. This statement, alone and by itself, is not sufficient.

Combined statements. With the two statements, we know both (M – K) and the difference of the squares, so we can solve for (M + K). Together, the statements are sufficient.

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In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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03 May 2015, 05:17
What are the correct values for M and K? Solved it in a different way and would like to know if my calculation was correct or if i just correctly guessed C. Much appreciated!
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In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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20 Sep 2016, 06:39
1. we can't get any info from here
2. we can find AE, AH, and side of the inner square - K. but we don't know for M.

1+2 - we know K, we can find M. sufficient.
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Re: In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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20 Sep 2016, 16:17
Statement 1 is insufficient because it does not give us anything relevant m-k = 9

Statement 2 is insufficient but is helpful in the following way:
We know the area of bigger square = m^2
Area of smaller square is K^2
We can infer that M^2 = K^2 + (4*36) {area of smaller square plus area of 4 triangles}

This gives us M^2 - K^2 = 144
(M-K)(M+K) = 144

To solve we can get M-K from first statement so C is sufficient
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Re: In the diagram above, ABCD is a square with side M, and EFGH is a squa  [#permalink]

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29 Jun 2019, 06:24
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: In the diagram above, ABCD is a square with side M, and EFGH is a squa   [#permalink] 29 Jun 2019, 06:24
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