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chetan2u
Attachment:
CIRCLE1.png
A, B, C and D are points on the circumference of circle with center at O. AC is the diameter of the circle and OD is parallel to BC. If the area of the sector formed by radius OD and OC, shown by blue colour, is \(\frac{1}{6}\)th of area of circle, what is the measure of line DE?

(A) \(\sqrt{3}\)
(B) \(\sqrt{2}\)
(C) 1
(D) \(\frac{\sqrt{3}}{2}\)
(E) \(\frac{1}{2}\)

New question
self made

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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triangles ODE and ABC are similar.
BC corresponds to DE
AC corresponds to OD

thus:

DE/BC equals OD/AC
DE/1 equals r/2r
DE equals 1/2.

where am i going wrong?
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Mansoor50
triangles ODE and ABC are similar.
BC corresponds to DE
AC corresponds to OD

thus:

DE/BC equals OD/AC [this part is wrong]
DE/1 equals r/2r
DE equals 1/2.

where am i going wrong?

\(\triangle\) 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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Mansoor50
triangles ODE and ABC are similar.
BC corresponds to DE
AC corresponds to OD

thus:

DE/BC equals OD/AC [this part is wrong]
DE/1 equals r/2r
DE equals 1/2.

where am i going wrong?

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

Thank u!
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