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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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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 [ 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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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 [ 3.75 KiB | Viewed 1232 times ]

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

$$Area = \frac{9\sqrt{3}}{4}$$

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:

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

$$Area = \frac{9\sqrt{3}}{4}$$

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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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19 Jan 2016, 19:34
1
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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10 Jul 2016, 02:08
Bunuel wrote:

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

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