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In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°.

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In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°.  [#permalink]

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New post 16 Jul 2018, 04:41
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In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°. FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF?
Attachment:
hexagon.PNG
hexagon.PNG [ 22.98 KiB | Viewed 544 times ]


A) \(\frac{1}{6}\)
B) \(\frac{1}{8}\)
C) \(\frac{1}{12}\)
D) \(\frac{1}{18}\)
E) \(\frac{1}{24}\)
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In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°.  [#permalink]

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New post 16 Jul 2018, 05:02
Bismarck wrote:
In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°. FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF?
Attachment:
The attachment hexagon.PNG is no longer available


A) \(\frac{1}{6}\)
B) \(\frac{1}{8}\)
C) \(\frac{1}{12}\)
D) \(\frac{1}{18}\)
E) \(\frac{1}{24}\)



pl mention source if known

two ways....
Pl refer to attached fig

1) AGB is a equilateral triangle and OB is dividing this equilateral triangle in two equal parts.
when we join all vertices with centre G, all will form such 6 equilateral triangles and each equilateral triangle is divided in 2 equal parts, thus 12 parts
\(\triangle {AOB}\) is one of these 12, so area of AOB is \(\frac{1}{12}\) of the total

2) Geometry
each angle of hexagon is 120, so AO divides the angle in two parts so \(\angle{BAO}=\frac{120}{2}=60\)
so \(\triangle {AOB}\) is 30:60:90 so sides OA:OB:AB = 1:\(\sqrt{3}\):2
let side of hexagon=AB=2x, so sides OA =x and OB=\(x*\sqrt{3}\)
Area of \(\triangle {AOB}=\frac{1}{2}*x*x\sqrt{3}=x^2\sqrt{3}/2\)
area of hexagon = 6* area of equilateral triangle AGB= \(6*\frac{sqrt3}{4}*(2x)^2=6\sqrt{3}x^2\)
\(\frac{area-of-AOB}{area-0f-hexagon}=x^2\sqrt{3}x^2/2*6x^2\sqrt{3}=\frac{1}{12}\)

C
Attachments

hexagon.PNG
hexagon.PNG [ 25.88 KiB | Viewed 488 times ]


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Re: In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°.  [#permalink]

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New post 16 Jul 2018, 05:29
Bismarck wrote:
In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°. FO is parallel to ED. What is the ratio of the area of the triangle AOF to that of the hexagon ABCDEF?
Attachment:
The attachment hexagon.PNG is no longer available


A) \(\frac{1}{6}\)
B) \(\frac{1}{8}\)
C) \(\frac{1}{12}\)
D) \(\frac{1}{18}\)
E) \(\frac{1}{24}\)


Please refer the affixed diagram,

Area of given Hexagon=12* Area of triangle AOF (All triangles have equal area)
So, desired ratio=\(\frac{1}{12}\)

Ans. (C)
Attachments

Hexagon 12.JPG
Hexagon 12.JPG [ 22.08 KiB | Viewed 502 times ]


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Re: In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°. &nbs [#permalink] 16 Jul 2018, 05:29
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In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°.

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