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

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Joined: 18 Jun 2018
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In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°.  [#permalink]

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16 Jul 2018, 04:41
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35% (medium)

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67% (02:14) correct 33% (04:11) wrong based on 18 sessions

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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 [ 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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Joined: 02 Aug 2009
Posts: 6792
In the figure below, ABCDEF is a regular hexagon and angle AOF = 90°.  [#permalink]

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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 [ 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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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) $$\frac{1}{24}$$

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 [ 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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