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In the preceding figure, circle O has its center at the origin. If E A

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In the preceding figure, circle O has its center at the origin. If E A  [#permalink]

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New post 17 Jan 2016, 11:56
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In the preceding figure, circle O has its center at the origin. If E AOP = 30°, what is the area of ∆OAB?

A. \(\sqrt{3}\)

B. \(\frac{9 \sqrt{3}}{4}\)

C. \(3 \sqrt{2}\)

D. \(3 \sqrt{3}\)

E. 9

Attachment:
2016-01-17_2253.png
2016-01-17_2253.png [ 7.5 KiB | Viewed 1415 times ]

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In the preceding figure, circle O has its center at the origin. If E A  [#permalink]

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New post 19 Jan 2016, 12:03
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To find the area of triangle OAB, we'll need the length of one side and the perpendicular height. Area = bh/2

We can take OB as the base, since we already know its length is 3. We will then need the height of the triangle, which will be the x coordinate of point A. We are told that the angle AOP is \(30^{\circ}\), so we can make the point C on the x-axis with the same x coordinate as point A, and form a 30-60-90 triangle AOC. The hypotenuse of \(\triangle\)AOC is 3 (radius of the circle), and therefore the length of OC is \(\frac{3\sqrt{3}}{2}\)

We get this from the ratio of side lengths of a 30-60-90 triangle \(1:\sqrt{3}:2\). So since the hypotenuse is 3, the length of the long leg is \(\frac{3\sqrt{3}}{2}\)

Attachment:
Circle Triangle.png
Circle Triangle.png [ 3.75 KiB | Viewed 1232 times ]


Now we have the base and the height, so we can calculate the area
\(Area = \frac{1}{2}*3*\frac{3\sqrt{3}}{2}\)

\(Area = \frac{9\sqrt{3}}{4}\)

Answer: B

An alternative method would be to consider \(\triangle\)OAB and see that it is an isosceles triangle with legs OA and OB = 3. We also have the angle at O = \(120^{\circ}\). If we cut the triangle in half with a line perpendicular to AB and passing through O, then we have two 30-60-90 triangles with hypotenuse = 3. Using the \(1:\sqrt{3}:2\) ratio, the length of the short side is \(\frac{3}{2}\), which is also the height of \(\triangle\)OAB, and the length of the long side is \(\frac{3\sqrt{3}}{2}\), which is half the base of \(\triangle\)OAB. So the height of \(\triangle\)OAB is \(\frac{3}{2}\) and the base is \(\frac{6\sqrt{3}}{2}\).

From there we can calculate the area to be:

\(Area = \frac{1}{2}*\frac{6\sqrt{3}}{2}*\frac{3}{2}\)

\(Area = \frac{9\sqrt{3}}{4}\)

Answer: B
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Re: In the preceding figure, circle O has its center at the origin. If E A  [#permalink]

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New post 19 Jan 2016, 19:34
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Since P(3,0) , Radios is 3.
So two sides AO=OB =3
Now AOP=30 is given so the angle between two adjuscent sides = 120
Area = 0.5.AO.OB.Sin120
=0.5x3x3x(root(3)/2)
=B
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In the preceding figure, circle O has its center at the origin. If E A  [#permalink]

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New post 10 Jul 2016, 02:08
Bunuel wrote:
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In the preceding figure, circle O has its center at the origin. If E AOP = 30°, what is the area of ∆OAB?

A. \(\sqrt{3}\)

B. \(\frac{9 \sqrt{3}}{4}\)

C. \(3 \sqrt{2}\)

D. \(3 \sqrt{3}\)

E. 9

Attachment:
The attachment 2016-01-17_2253.png is no longer available



Attachment:
Circle Triangle.png
Circle Triangle.png [ 8.99 KiB | Viewed 984 times ]

Required area is ∆ODB+∆ODA
from below fig. angle AOB=120
OBA=OAB=30 as radius=3(isosceles ∆)
In ∆ DOB thus angleODB=60 Deg.
Now it is 30-60-90 ∆
So length OD=\(\sqrt{3}\)
Draw AC perpendicular height for ∆ODA
Angle OAC=60Deg.
Now ∆ ODA it is 30-60-90 triangle
So height AC=3/2
area is ∆ODB+∆ODA=1/2*3/2*\(\sqrt{3}\)+1/2*3*\(\sqrt{3}\)
=9*\(\sqrt{3}\)/4
Ans B
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Re: In the preceding figure, circle O has its center at the origin. If E A  [#permalink]

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Re: In the preceding figure, circle O has its center at the origin. If E A   [#permalink] 25 Jul 2017, 20:58
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