When opened and lying flat, a birthday card is in the shape of a regul

Solution:



When a regular hexagon is folded in half along one of its diagonals, it becomes an isosceles trapezoid which consists of 3 congruent regular triangles whose side length is also the side length of the hexagon (see diagram above). As a trapezoid, the shorter base of the trapezoid is a side of the hexagon, the longer base is the diagonal (along which the hexagon is folded). Furthermore, the longer base is exactly twice the length of the shorter base. Finally, since the nonparallel sides (i.e., the two sides that are not the bases) of the trapezoid have the length of a side of the hexagon, the height of the trapezoid, however long it is, is shorter than a side of the hexagon.

Statement One Only:

Each side of the regular hexagon is 4 inches long.

Since each side of the regular hexagon is 4 inches long, the shorter base of the trapezoid is 4 inches, the longer base is 2 x 4 = 8 inches, and the height is less than 4 inches. Therefore, it can fit into a rectangular envelope that is 4 inches by 9 inches. Statement one alone is sufficient.

Statement Two Only:

The area of the top surface (which is the same as the area of the bottom surface) of the folded birthday card is less than 36 square inches.

Since the folded birthday card has an area of 3 congruent equilateral triangles and if the card has an area of exactly 36 square inches, the area of each equilateral triangle will then be 12 square inches. Recall that the area of an equilateral triangle with side length of s is s^2√3/4. We can create an equation:

s^2√3/4 = 12

s^2 = 48/√3

s = √(48/√3) ≈ 5.26

However, since the area of the birthday card is less than 36 square inches, a side of the equilateral triangle (or a side of the regular hexagon) must be less than 5.26 square inches. If it’s 5 inches, then it won’t fit into an envelope that is 4 inches by 9 inches, since the longer base of the card (recall that it’s in the shape of a trapezoid) will then be 10 inches. On the other hand, if it’s 4 inches, then it will fit into an envelope that is 4 inches by 9 inches (see the analysis of statement one). Therefore, statement two alone is not sufficient to answer the question.

Answer: A
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IF WE USE THE BASIC CONCEPTS THIS PROBLEM CAN BE DONE IN SECONDS .

INFO GIVEN - the envelope has B=4 AND L=9.
The hexagon has been folded into half along one of its longest diagonal . note longest diagonal of hexagon is 2(side)

1)we know the side of Hexagon which gives us longest side of folded shape / longest diagonal of hexagon as 2*4 = 8 .
and we know the longest side of rectangular envelope is 9 . so we can easily fit the card into envelope .
SUFFICIENT .

2)Please note just by having same area of two different objects doesn't mean it will fit into each other . The shape matters (the sides matter ). NOT SUFFICIENT

won't statement 2 give us the sizes of the card?

I think given the area you can decide the length of the sides, like a right triangle.

You can actually solve this question by using very little math and by using some logic:

If you fold a hexagon from one of it's diagonal, you end up with a trapezoid with three sides as X (because a regular hexagon has equal sides), and Base 2 as Y. Now you need to know the values of X or Y to determine the dimensions of the figure. (Refer to pic attachment for a visual)

1.) Gives you the value of X, so you have enough information to determine a specific value of Y which will lead to your answer! SUFFICIENT

2.) The area is less than 36 square inches, but we don't know the values of either x or y. You can have a trapezoid that is super long but very small in height (assuming the area of it is 35.999999) or a trapezoid that's super tall but the base is super small. Or have a very small area (0.000001) and the right dimensions would be enough for the folded hexagon to fit in the envelope. NOT SUFFICIENT

Answer is A

The actual issue with this problem is at the core of data sufficiency.

You are mostly led to believe that statement II is sufficient because you have it in your head that each side is still 4. Given that the side is 4 and the area is 36 sure it is sufficient. But you have to get statement I completely out of your head and realize that there are a number of ways in which base 2 is in fact more than 9 yet still leads to an area of 36.

As a matter of fact you can calculate the area without being told the area just by knowing that side and chopping it up - but that's a waste of time for a problem like this. Simple logic will serve you best here.

Forget trying to prove why statement II doesn't work - just imagine statement I and II are flipped around and do the problem again and you will see that just having an area of 36 is not sufficient.

Bunuel can you help out with this please?

When hexagon is folded, it forms an isosceles trapezium as shown in the attached picture (Credit: jsound996)
Let's say the smaller parallel side is B1, longer parallel side is B2 and Height of trapezium is H.


Since regular hexagon divides into 6 equilateral triangles, we know:
1) B2 = 2B1
2) Height of equilateral triangle = √3*side/2

NEED: To make sure that we can satisfy 2 conditions: 1) H <=4 and 2) B2 <= 9.

1) B1 = 4; B2 = 8
Condition 2 met.
Let's calculate H.

Since regular hexagon divides into 6 equilateral triangles, we can calculate the height.
Height of equilateral triangle = √3*side/2 = 2√3 = 2 * 1.73 = 3.4 => Condition 1 met.

Since both conditions met, statement 1 is sufficient.

2) Area of trapezium = 0.5 (B1 + B2) * H
Statement says: Area < 36
0.5 (B1 + B2) * H < 36
0.5 (3B1) * H < 36

H = √3*side/2 = √3*B1/2

0.5 * (3B1) * √3*B1/2 < 36
(B1^2)*3√3/4 < 36
B1^2 < (4*36)/3√3
B1^2 < 4*12/√3
B1^2 < 4*4*3/√3
B1^2 < 4*4*√3
B1 < 4*3(1/4)
B1 < 4 * 1.3
B1 < 5.26

B2 < 2B1 => B2 < 10.52

Hence B2 could be less than or more than 9. Condition 2 not met.

Not sufficient.

ANSWER: A

Bunuel VeritasKarishma I was wondering whether there is a shorter way to do this Q. Thanks!

For stmnt 1, you don't need to calculate anything. Since it is a regular hexagon, knowing a side will give all other measurements so you will be able to say whether the card will fit in the envelope.
For stmnt 2, you will need to find the maximum value of the side to see if it may or may not fit in the envelope.
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VeritasKarishma

Hello Karishma,

The question says it is a regular hexagon.

The topside of the hexagon will have 3 equilateral triangles.
Let it side be s.
Then according to the question Area of a top hexagon is less than 36

Solving it we get s<24 * root 3

Therefore insufficient as we can have both yes and no case here.
Is this approach ok?

Certainly you can use the concept of equilateral triangles. The diagonal along the fold will be made up of two sides of equilateral triangles so its length will be 2s.
You will get that s < 5.something
So 2s < 10.something

Hence not sufficient since 2s may be less than 9 or may not be.
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Thank you for a wonderful explanation!!

Sure thing!!
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Hello ScottTargetTestPrep,
A prompt question regarding this problem, shouldn't we also do some calculations to see if there is a possibility that the folded hexagon could fit by having its base the diagonal of the envelope (because the diagonal is going to be greater that the base) I checked that in this case even the diagonal will not be sufficient, but I'm curious because the majority of posts takes as the greatest dimension the 9inch whereas the diagonal is about 9,85

Response:

Good question. It is clear that if we try to fit the folded hexagon in such a way that the base of the folded hexagon coincides with the diagonal of the envelope, it won’t fit. A diagonal of the rectangle divides the rectangle into two congruent right triangles, each with an area of 36/2 = 18. As soon as we choose a folded hexagon with an area greater than 18 square inches, it won’t fit. Thus, there are folded hexagons with areas less than 36 square inches that won’t fit into the rectangle that way.

If we improve your suggestion and consider the cases where the base of the hexagon is not necessarily the diagonal of the rectangle, but it can be any line segment that fits in the rectangle, it still won’t work. We are told that the area of the folded hexagon is less than 36 square inches. While this area cannot equal 36, it can get very close to 36; for example, the area of the folded hexagon can be 35.99. Since the area of the 4 by 9 rectangle is exactly 36 square inches, if a folded hexagon with an area close to 36 square inches fits into the rectangle, there must be very little area within the rectangle that is not occupied by the folded hexagon. Considering all rectangles that could contain the folded hexagon which is in the shape of an isosceles trapezoid, we see that there must always be plenty of space between the rectangle and the trapezoid. Thus, we can safely conclude that there are folded hexagons with areas less than 36 square inches which will not fit into a 4 by 9 rectangle.
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Why have we considered only the longest diagonal while attempting to fold the card? What if we fold it across the shorter diagonal?

Response:

The question tells us that the birthday card must be “folded in half.” If we fold the card along one of its shorter diagonals, then the birthday card is not folded in half. It is clear that the question is talking about one of the longer diagonals.
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For statement 2:

area of a regular hexagon = \(\frac{s^23\sqrt{3}}{2}\)

72 = \(\frac{s^23\sqrt{3}}{2}\)

48/\(\sqrt{3}\) = s^2

4*~1,3 = s

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