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A chord AB of a circle subtends an angle of 120° at the centre O

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QuantMadeEasy
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A chord AB of a circle subtends an angle of 120° at the centre O [#permalink] New post   14 Jul 2020, 01:53
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A chord AB of a circle subtends an angle of 120° at the centre O. If the radius of the circle is 2. What is the area of the triangle AOB?​

A. 1​

B. √3/2​

C. √2​

D. √3​

E. Cannot be determined​
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D

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A chord AB of a circle subtends an angle of 120° at the centre O [#permalink] New post   14 Jul 2020, 02:41
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IMO D

The chord creates an isosceles triangle. With OA = OB = 2

A perpendicular to the chord from AB (Height of triangle) makes two 30 - 60 - 90 triangle with 90 degree angle opposite to one of the sides .

The sides of triangle are (a : a * Sqrt(3) :2a) >> (1 : 1 *Sqrt(3) :2)

So length of chord AB = 2* Sqrt(3)

Area of triangle = 1/2 * 2*Sqrt(3) * 1 ====> sqrt (3)

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Re: A chord AB of a circle subtends an angle of 120° at the centre O [#permalink] New post   15 Jul 2020, 03:18
Using ab sin c I am getting 2*2*√3/2 =√3*2

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Re: A chord AB of a circle subtends an angle of 120° at the centre O [#permalink] New post   15 Jul 2020, 06:22
shramana wrote:
Using ab sin c I am getting 2*2*√3/2 =√3*2

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Area formula is 1/2*A*B*SinC
which will give area as = 1/2*2*2*Sin 120
1/2*2*2*√3/2
=> √3
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A chord AB of a circle subtends an angle of 120° at the centre O [#permalink] New post   15 Jul 2020, 08:46
QuantMadeEasy wrote:
A chord AB of a circle subtends an angle of 120° at the centre O. If the radius of the circle is 2. What is the area of the triangle AOB?​

A. 1​

B. √3/2​

C. √2​

D. √3​

E. Cannot be determined​



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Screenshot 2020-07-15 at 9.15.22 PM.png
Screenshot 2020-07-15 at 9.15.22 PM.png [ 23.95 KiB | Viewed 2920 times ]


Area of\( \triangle \)AOB = 1/2* AB * CO = \(1/2 * 2\sqrt{3}*1 = \sqrt{3}\)

IMO D
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A chord AB of a circle subtends an angle of 120° at the centre O [#permalink]
15 Jul 2020, 08:46
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