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17 Jan 2016, 11:56
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
74% (03:01) correct
26%
(02:50)
wrong
based on 90
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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 3583 times ]
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 \(\frac{3\sqrt{3}}{2}\)
Circle Triangle.png [ 3.75 KiB | Viewed 3325 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
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 3325 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
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
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
