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05 Oct 2021, 01:30
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
70% (01:34) correct
30%
(01:44)
wrong
based on 77
sessions
History
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Time
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Not Attempted Yet
In the given figure, O is the centre of the circle of radius 1 unit. What is the area, in square units, of the minor sector AOC?
A \(\frac{π}{6}\)
B \(\frac{π}{3}\)
C \(\frac{π}{2}\)
D \(\frac{2π}{3}\)
E \(π\)
Attachment:
1.jpg [ 9.79 KiB | Viewed 6602 times ]
05 Oct 2021, 02:08
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Area of circle: \(πr^2\)
\(π(1)^2\)
\(π\)
\(OA = OC = OB\) as they are all radii.
Drawing in line OB will therefore create 2 isosceles triangles. This also means that angles \(OCB = OAB = OBC = OBA\).
Which means that \(angle ABC = 60'\), and because of Central Angle Theorem \(angle AOC = 120'\).
As \(AOC\) is 120', it means that \(arc AOC\) is \(\frac{ 1}{3}\) of the circle.
Thus the area of minor sector \( AOC = \frac{π}{3}\)
Answer B
\(π(1)^2\)
\(π\)
\(OA = OC = OB\) as they are all radii.
Drawing in line OB will therefore create 2 isosceles triangles. This also means that angles \(OCB = OAB = OBC = OBA\).
Which means that \(angle ABC = 60'\), and because of Central Angle Theorem \(angle AOC = 120'\).
As \(AOC\) is 120', it means that \(arc AOC\) is \(\frac{ 1}{3}\) of the circle.
Thus the area of minor sector \( AOC = \frac{π}{3}\)
Answer B
05 Oct 2021, 06:26
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Bunuel
In the given figure, O is the centre of the circle of radius 1 unit. What is the area, in square units, of the minor sector AOC?
A \(\frac{π}{6}\)
B \(\frac{π}{3}\)
C \(\frac{π}{2}\)
D \(\frac{2π}{3}\)
E \(π\)
Attachment:
1.jpg
Total angle of a circle =360
=>120 +120 +x = 360
=>x= 120
Area of the sector =x/360 * pie * 1^2
=>1/3 * pie
Therefore IMO B
