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In the figure above, triangle ABC is inscribed in a circle whose centr

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Bunuel
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In the figure above, triangle ABC is inscribed in a circle whose centr [#permalink] New post   03 Jun 2020, 08:05
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In the figure above, triangle ABC is inscribed in a circle whose centre O has the x- and y-coordinates as (0,0). If the x- and y- coordinates of point A are (-4,0) and ∠BAC = 30∘, what is the area, in square units, of triangle AOB?

A. 2√3
B. 4√3
C. 8√3
D. 16√3
E. 24√3

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Re: In the figure above, triangle ABC is inscribed in a circle whose centr [#permalink] New post   03 Jun 2020, 09:17
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Area of triangle ABC = 1/2× 4 × 4sqrt3
= 8sqrt3
Area of triangle OBC= sqrt3× (4)^2÷4
= 4sqrt3
Area of triangle AOB = 8sqrt3-4sqrt3=4sqrt3
Option B is the answer

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Re: In the figure above, triangle ABC is inscribed in a circle whose centr [#permalink] New post   03 Jun 2020, 10:03
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Angle subtends by chord at the centre is double the angle subtended by chord at the circle.
Therefore 2 ⟨BAC = ⟨BOC
=> 2*30° = Angle BOC => 60°
OB = OC (radius)
Interior angle of a triangle=
Angle BOC+CBO+BCO = 180
CBO+BCO = 120°
2CBO = 120° => CBO = 60°
Triangle CBO is an equilateral triangle.

Radius = 4
Distance between OA
= √(y2-y1)^2+(x2-x1)^2
= √(0-0)^2+(0+4)^2
= √4^2 => 4
OA=OB (radius)

Area of Triangle CBO is √3/4r^2
√3*4*4/4 => 4√3

Answer is B

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Re: In the figure above, triangle ABC is inscribed in a circle whose centr [#permalink] New post   13 Jul 2021, 07:18
Angle subtended at the cenre of semicircle is equal to 90 therefore <aob = 90
ob=oa = radius therefore <abc = 30
which provides <aob = 120 and <cob = <cbo= 60
therefore triangle cbo is equilateral
required area = tri abc- tri cbo
area of tri abc = 1/2× 4 × 4sqrt3= 8sqrt3
area of tri obc= sqrt3× (4)^2÷4= 4sqrt3
Area of tri aob = 8sqrt3-4sqrt3=4sqrt3
therefore IMO B
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Re: In the figure above, triangle ABC is inscribed in a circle whose centr [#permalink] New post   14 Jul 2021, 05:39
Radius of the circle- 4
Diameter of the circle - 8

Angle ABC - 90 (Angle in a semicircle from it's Diameter is 90)

BAC = 30
ABC = 90
BCA = 180 - 120 = 60

Apply 30 60 90 rule

AB = 4 root 3
AO = 4

Area AOB = 1/2×4×4rt3×sin 30
Area = 4rt3

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Re: In the figure above, triangle ABC is inscribed in a circle whose centr [#permalink]
14 Jul 2021, 05:39
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