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21 Sep 2019, 18:21
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A
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57% (02:36) correct
43%
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wrong
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When opened and lying flat, a birthday card is in the shape of a regular hexagon. The card must be folded in half along 1 of its diagonals before being placed in an envelope for mailing. Assuming that the thickness of the folded card will not be an issue, will the birthday card fit inside a rectangular envelope that is 4 inches by 9 inches?
(1) Each side of the regular hexagon is 4 inches long.
(2) 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.
DS61791.01
(1) Each side of the regular hexagon is 4 inches long.
(2) 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.
DS61791.01
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26 Jul 2020, 15:43
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gmatt1476
Solution:
Attachment:
hex birthday card.png [ 3.02 KiB | Viewed 20010 times ]
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
01 Oct 2019, 23:37
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gmatt1476
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
