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03 Jun 2020, 04:46
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C
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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
Attachment:
Q.png [ 10.65 KiB | Viewed 3716 times ]
yashikaaggarwal
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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
Posted from my mobile device
‹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
Posted from my mobile device
General Discussion
03 Jun 2020, 06:25
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Kindly see the attachment
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1.PNG [ 78.6 KiB | Viewed 3386 times ]
