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In the given figure, AB and CD are the longest chords of the circle wi

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Bunuel
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In the given figure, AB and CD are the longest chords of the circle wi [#permalink] New post  03 Jun 2020, 04:46
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In the given figure, AB and CD are the longest chords of the circle with their lengths equal to 8 units. If the length of the minor arc BC is 1/6th of the perimeter of the circle, what is the length of the chord BD?

A. 2√3
B. 4
C. 4√3
D. 8
E. 8√3


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yashikaaggarwal
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Re: In the given figure, AB and CD are the longest chords of the circle wi [#permalink] New post  03 Jun 2020, 05:07
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Arc BC is 1/6 of whole circle. That means angle BOC subtends 60° angle at the centre.
‹BOC=‹AOD (opposite angles)
Join AD.
In triangle AOD.
Angle AOD is 60°
AO = DO (radius) hence AOD is an equilateral triangle
AO=DO=AD

Since, AB is the biggest Chord.
It subtends 90°angle on the circle, therefore ∆ABD is a right angle triangle. In which,
AB=8 & AD=4
Also AB^2= AD^2+BD^2
8^2=4^2+BD^2
64-16=BD^2
48=BD^2
BD=4√3

Answer is C

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Re: In the given figure, AB and CD are the longest chords of the circle wi [#permalink] New post  03 Jun 2020, 06:25
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satya2029
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In the given figure, AB and CD are the longest chords of the circle wi [#permalink] New post  03 Jun 2020, 07:11
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Bunuel wrote:

In the given figure, AB and CD are the longest chords of the circle with their lengths equal to 8 units. If the length of the minor arc BC is 1/6th of the perimeter of the circle, what is the length of the chord BD?

A. 2√3
B. 4
C. 4√3
D. 8
E. 8√3


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Longest chord= Diameter= 2* radius
Radius= 4
perimeter= 8pi
Since CBD is half of the perimeter, and we know BC in terms of perimeter
4pi=1/6*8pi+arc BD
Arc BD= 4pi-4pi/3=8pi/3
and the angle that the arc BD makes with the center is
\((360/8pi)* 8pi/3\)
120
let the center is O so in triangle BOD
it's 30-30-120 as OB=OD and the sides are in the ratio of \(1:1:1\sqrt{3}\)
OB=OD=4
Hence \(BD=4\sqrt{3}\)
C:)

Or
as BC is 1/6th of the perimeter, it makes 60 degree with the center so in BOD, angle at O is 120
it's 30-30-120 as OB=OD and the sides are in the ratio of \(1:1:1\sqrt{3}\)
OB=OD=4
Hence \(BD=4\sqrt{3}\)
C:)
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Re: In the given figure, AB and CD are the longest chords of the circle wi [#permalink] New post  03 Jun 2020, 09:43
Bunuel wrote:

In the given figure, AB and CD are the longest chords of the circle with their lengths equal to 8 units. If the length of the minor arc BC is 1/6th of the perimeter of the circle, what is the length of the chord BD?

A. 2√3
B. 4
C. 4√3
D. 8
E. 8√3



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IMO C . Solution in the figure attached .
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Re: In the given figure, AB and CD are the longest chords of the circle wi [#permalink] New post  03 Jun 2020, 09:57
The chords are longest implies AB and CD are diameters.
=> CBD form a right triangle with CBD=90.
Let AB and CD intersect at O. Since BD is \(\frac{1}{6}\)th of the circumference, therefore COB=60. => OBD = ODB =30.
Thus, CBD is a 30-60-90 triangle and it's hypotenuse is 8 units.
So, BD = 4\(\sqrt{3}\)
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NitishJain
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In the given figure, AB and CD are the longest chords of the circle wi [#permalink] New post  03 Jun 2020, 10:52
IMO C.

Greatest chords of the circle = diameter of the circle

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Re: In the given figure, AB and CD are the longest chords of the circle wi [#permalink] New post  18 Sep 2021, 08:48
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