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16 Jul 2018, 04:41
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (02:04) correct
26%
(02:55)
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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?
hexagon.PNG [ 22.98 KiB | Viewed 25487 times ]
A) \(\frac{1}{6}\)
B) \(\frac{1}{8}\)
C) \(\frac{1}{12}\)
D) \(\frac{1}{18}\)
E) \(\frac{1}{24}\)
Attachment:
hexagon.PNG [ 22.98 KiB | Viewed 25487 times ]
A) \(\frac{1}{6}\)
B) \(\frac{1}{8}\)
C) \(\frac{1}{12}\)
D) \(\frac{1}{18}\)
E) \(\frac{1}{24}\)
16 Jul 2018, 05:02
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Bismarck
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 25067 times ]
16 Jul 2018, 05:29
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Bismarck
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 [ 22.08 KiB | Viewed 24251 times ]
