syedmohammad211 wrote:
Bunuel can you help out with this please?
gmatt1476 wrote:
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
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!
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