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Point O is the center of the semicircle. If ∠BCO = 30° and BC = 6√3

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Point O is the center of the semicircle. If ∠BCO = 30° and BC = 6√3  [#permalink]

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New post 24 Jan 2018, 07:17
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Point O is the center of the semicircle. If ∠BCO = 30° and BC = 6√3, what is the area of triangle ABO?

A. 4√3
B. 6√3
C. 9√3
D. 12√3
E. 24√3

Attachment:
2018-01-24_1916.png
2018-01-24_1916.png [ 12.41 KiB | Viewed 931 times ]

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Re: Point O is the center of the semicircle. If ∠BCO = 30° and BC = 6√3  [#permalink]

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New post 24 Jan 2018, 07:25
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From the diagram we can conclude that the Triangle ABC is a right triangle with 30-60-90 having sides in the ratio 1:rt(3):2.
Hence the dia AC= 12 and AB=6.

in triangle AOB we have OA=AB=6. also angle A is 60deg. Hence triangle AOB is an equilateral triangle with side 6. Height of this triangle will be 3*rt(3). Hence area of AOB is 1/2*3*rt(3)*6=9*rt(3)
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Re: Point O is the center of the semicircle. If ∠BCO = 30° and BC = 6√3  [#permalink]

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New post 24 Jan 2018, 08:51
Bunuel wrote:
Image
Point O is the center of the semicircle. If ∠BCO = 30° and BC = 6√3, what is the area of triangle ABO?

A. 4√3
B. 6√3
C. 9√3
D. 12√3
E. 24√3

Attachment:
2018-01-24_1916.png


From, the figure \(Angle OCB = Angle OBC = 30, Angle BOC = 120.\) Henace, \(Angle OBA = Angle OAB = Angle BOA = 60\).. This is now a Equilateral Traingle. Also, \(Angle ABC = 90. AB = 6.\)
\(Area = 9\sqrt{3}\)
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Point O is the center of the semicircle. If ∠BCO = 30° and BC = 6√3  [#permalink]

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New post 24 Jan 2018, 21:54
Bunuel wrote:
Image
Point O is the center of the semicircle. If ∠BCO = 30° and BC = 6√3, what is the area of triangle ABO?

A. 4√3
B. 6√3
C. 9√3
D. 12√3
E. 24√3

Attachment:
The attachment 2018-01-24_1916.png is no longer available

Attachment:
inscribedtri.png
inscribedtri.png [ 60.39 KiB | Viewed 638 times ]

Rule: If A, B, and C are distinct points on a circle where the line AC is the diameter of the circle, then \(\angle\) ABC is a right angle.
OR: If the diameter of a circle is an inscribed triangle's hypotenuse, the triangle is a right triangle.
AC is the diameter of the circle and the hypotenuse of inscribed \(\triangle\) ABC
\(\triangle\) ABC is a right triangle

1) \(\triangle\) ABC = 30-60-90 triangle
Right \(\angle\) ABC = 90°. Given, ∠BCO = 30°. Hence ∠BAO = 60°

30-60-90 triangles have side lengths that correspond to 30-60-90, in ratio \(x: x\sqrt{3}: 2x\)
Side BC = 6√3. Opposite the 60° angle, 6√3 corresponds to x√3
Thus x = 6, side AB = x = 6, and side AC = 2x = 12
Side AC is the circle's diameter, d
d = 12 = 2r, r = 6

2) \(\triangle\) ABO side lengths
Sides AO and OB of \(\triangle\) ABO are both radii, with length = 6
Side AB = 6 (from above)
\(\triangle\) ABO's sides all = 6. It is equilateral.

3) AREA of equilateral \(\triangle\) ABO*

Area of an equilateral triangle is \(\frac{s^2\sqrt{3}}{4}\)
\(\frac{6^2\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} = 9\sqrt{3}\)

Answer C

*If you don't remember the formula for the area of an equilateral triangle, draw one. Drop an altitude, which is a perpendicular bisector of the opposite side and of the vertex.
That altitude creates two congruent right 30-60-90 triangles, as above. Side lengths correspond to 30-60-90, in ratio \(x : x\sqrt{3} : 2x\)

Side lengths? Side opposite the 90° angle = \(6 = 2x\). So side opposite 30° angle is half of that, i.e., \(x, x = 3\). Side opposite 60° angle = height of triangle = \(x\sqrt{3}\) or \(3\sqrt{3}\).

Area, where base AO = 6: \(\frac{b*h}{2} = (6*3\sqrt{3})*\frac{1}{2} = 9\sqrt{3}\)

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Point O is the center of the semicircle. If ∠BCO = 30° and BC = 6√3 &nbs [#permalink] 24 Jan 2018, 21:54
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