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A, B, C and D are points on the circumference of circle with center at

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A, B, C and D are points on the circumference of circle with center at  [#permalink]

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New post Updated on: 23 Jul 2018, 23:40
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A
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C
D
E

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Question Stats:

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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


Attachment:
CIRCLE1.png
CIRCLE1.png [ 4.82 KiB | Viewed 1035 times ]

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Originally posted by chetan2u on 23 Jul 2018, 22:10.
Last edited by Bunuel on 23 Jul 2018, 23:40, edited 1 time in total.
Edited the question.
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Re: A, B, C and D are points on the circumference of circle with center at  [#permalink]

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New post 23 Jul 2018, 22:39
1
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
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Re: A, B, C and D are points on the circumference of circle with center at  [#permalink]

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New post 23 Jul 2018, 22:43
dprchem wrote:
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


Yes, you are correct. it is all about 30:60:90 triangle and parallel sides..
you can also solve without any further drawing..
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: A, B, C and D are points on the circumference of circle with center at  [#permalink]

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New post 23 Jul 2018, 23:38
chetan2u wrote:
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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Re: A, B, C and D are points on the circumference of circle with center at  [#permalink]

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New post 27 Jul 2018, 17:31
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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Re: A, B, C and D are points on the circumference of circle with center at  [#permalink]

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New post 27 Jul 2018, 21:08
Mansoor50 wrote:
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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Re: A, B, C and D are points on the circumference of circle with center at  [#permalink]

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New post 29 Jul 2018, 02:21
Princ wrote:
Mansoor50 wrote:
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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Re: A, B, C and D are points on the circumference of circle with center at &nbs [#permalink] 29 Jul 2018, 02:21
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