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In the figure above, ABCDEF is a regular hexagon. If area of ACE is 1
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19 Jun 2023, 05:38
19 Jun 2023, 05:38
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In the figure above, ABCDEF is a regular hexagon. If area of ΔACE is \(100 \sqrt{3} cm^2\), what is the area of the hexagon?
A. \(150 \sqrt{3}\)
B. \(160 \sqrt{3}\)
C. \(180 \sqrt{3 }\)
D. \(200 \sqrt{3}\)
E. \(240 \sqrt{3}\)
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Re: In the figure above, ABCDEF is a regular hexagon. If area of ACE is 1
[#permalink]
16 Jul 2023, 04:40
16 Jul 2023, 04:40
Expert Reply
OE
We know that a regular hexagon with side s can be divided into six equilateral triangles
Here, if we draw line segments from A, C and E respectively, they meet at O and has a
measure equal to the side of the hexagon, s.
At O, the degree measure is 360o
.
The segments AO, CO and EO divides the angle in three equal measures that is, 120o
.
Also, in ∆AOC, AO = OC which gives that ∆AOC is a 30-30-120 triangle.
Similarly, in ∆ABC, AB = BC and ∠ABC = 120°. (The degree measure of each angle in a regular hexagon is 120°)
Hence, ∆ABC is also a 30-30-120 triangle.
In ∆AOC and ∆ABC, AC is common, and all angles are also equal.
Hence the area of both triangles is also equal. Let the area be X.
Similarly, we can divide the hexagon into six triangles with area X
Now, we are given that area of ∆ACE = 100√3 that is, 3X = 100√3
Area of the hexagon, ABCDEF = 6X
a. = 2 (3X)
b. = 2 (100√3)
c. = 200√3
D is the answer
We know that a regular hexagon with side s can be divided into six equilateral triangles
Here, if we draw line segments from A, C and E respectively, they meet at O and has a
measure equal to the side of the hexagon, s.
At O, the degree measure is 360o
.
The segments AO, CO and EO divides the angle in three equal measures that is, 120o
.
Also, in ∆AOC, AO = OC which gives that ∆AOC is a 30-30-120 triangle.
Similarly, in ∆ABC, AB = BC and ∠ABC = 120°. (The degree measure of each angle in a regular hexagon is 120°)
Hence, ∆ABC is also a 30-30-120 triangle.
In ∆AOC and ∆ABC, AC is common, and all angles are also equal.
Hence the area of both triangles is also equal. Let the area be X.
Similarly, we can divide the hexagon into six triangles with area X
Now, we are given that area of ∆ACE = 100√3 that is, 3X = 100√3
Area of the hexagon, ABCDEF = 6X
a. = 2 (3X)
b. = 2 (100√3)
c. = 200√3
D is the answer
gmatclubot
Re: In the figure above, ABCDEF is a regular hexagon. If area of ACE is 1 [#permalink]
16 Jul 2023, 04:40
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