Hi,

What is the answer to this question?

I calculated as below:

Angle DOC = 60 since sector DOC is 1/6 of area of circle
Angle ABC =90 since angle made by diameter
Now extend DO to meet AB at F
Angle AOF=DOC=60
Since OD is parallel to BC, OF is also parallel to BC
Then triangle AOF and ABC are similar triangles and hence both are 30-60-90 triangles.
From this it can be deduced that AC=2 or radius =1 (calculated using BC=1)
Hence OD=1
Now Triangle ODE is also a 30-60-90 Triangle
Hence DE =sqrt(3)/2
Answer: D

OA: D
\(\angle COD = 60^{\circ}\) as colored sector area is \(\frac{1}{6}\) of total circle's area
\(\angle COD = \angle BCA = 60^{\circ}\) (Alternate Opposite angles, \(BC||OD\))

Both \(\triangle ACB\) and \(\triangle DOE\) are \(30^{\circ}-60^{\circ}-90^{\circ} \triangle\)

In \(\triangle ACB\), Ratio of sides \(BC:AB:AC\) is \(1:\sqrt{3}:2\).

As\(BC =1\), then \(AC\) would be\(2\).

\(AC = 2 OD\)(Radius of circle), \(OD = \frac{AC}{2} =\frac{2}{2} =1\)

In \(\triangle DOE\),Ratio of sides \(OE:ED:OD\) is \(1:\sqrt{3}:2\).

As \(OD =1\), \(ED\) would be \(\frac{\sqrt{3}}{2}\)
_________________

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\(\triangle\) ODE and CAB are similar.
\(\frac{OD}{AC}(Opposite \angle=90^{\circ})=\frac{DE}{AB}(Opposite \angle =60^{\circ})=\frac{OE}{BC}(Opposite \angle = 30^{\circ})\)
_________________

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