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Inscribed Triangle (tricky)

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Inscribed Triangle (tricky)

by El Cucu » Sat Jun 20, 2009 4:52 pm
For the triangle shown above, where A, B and C are all points on a circle, and line segment AB has length 18, what is the area of triangle ABC?

(1) Angle ABC measures 30°.

(2) The circumference of the circle is 18pi .
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by SanjeevK » Sat Jun 20, 2009 5:03 pm
IMO [spoiler]E:[/spoiler]
A: This doesn't provide any information. AB can be diameter or it can be a chord. The area of the triangle can't be calculated
B: This gives the radius of the circle. But this doesn't help in calculating the area

Taking A and B together also doesn't provide much information.

What is the OA?
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by El Cucu » Sat Jun 20, 2009 6:09 pm
SanjeevK wrote:IMO [spoiler]E:[/spoiler]
A: This doesn't provide any information. AB can be diameter or it can be a chord. The area of the triangle can't be calculated
B: This gives the radius of the circle. But this doesn't help in calculating the area

Taking A and B together also doesn't provide much information.

What is the OA?
OA is C. I found this problem in Manhat. Cat 5.
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by rahulg83 » Sun Jun 21, 2009 1:49 am
I got C
A and B alone are insufficient as SanjeevK pointed out.
but from two we can say that radius = 9/pi, or diameter = 18/pi

that means somewhere inside the triangle, we have the center. (How?)name it O. Noe join AO and CO. angle AOC = 60 and triangle AOC is an equilateral triangle as angle OAC = angle OCA = (180-60)/2 = 60. Therefore length of AC = radius of circle = 9/pi. Now we have two sides of a triangle and one known angle, which is opposite to one of the sides, we can calculate the remaining angles, remaining sides, and also the area of triangle.

Use formula sin (angle A)/BC = sin (angle B)/CA = sin (angle C)/AB

Heron's formula which gives the area in terms of the three sides of the triangle, specifically, as the square root of the product s(s – a)(s – b)(s – c) where s is the semiperimeter of the triangle, that is, s = (a + b + c)/2.

P.S. its better to try not to understand like this, try to make the circle and triangle in a piece of paper.
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by barnum » Sun Jun 21, 2009 8:04 am
A little bit easier to understand is that statement 2 tells you that the circ is 18pi (I assume you just left the pi out of your original post). This means that the diameter is 18, which is shown in the picture as the long side of the triangle. This means that that long side is the diameter. Any triangle drawn inscribed in a circle whose long side is the diameter will always be a right triangle. That means you now know that ACB is a 90 degree angle.

Taken together with statement 1 you now know it is a 30-60-90 triangle, which means you can use the ratio 1:sgrt3:2 to figure out the legs of the triangle. Since it is a right triangle, the legs can be used to calculate the area of the triangle.

If you need more explanation on how to do any of those steps, let me know.
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by iamcste » Sun Jun 21, 2009 8:36 am
barnum wrote:A little bit easier to understand is that statement 2 tells you that the circ is 18pi (I assume you just left the pi out of your original post).
Can the Original poster confirm whether the circumference is 18pi or just 18

Either way, you get C but if its 18pi, perhaps problem becomes easier.
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by barnum » Sun Jun 21, 2009 9:09 am
It would have to be 18pi. If it were 18, then the diameter would be less than 6. This means that you couldn't possibly have a line that was 18 long in the circle.
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by iamcste » Sun Jun 21, 2009 10:16 am
barnum wrote:It would have to be 18pi. If it were 18, then the diameter would be less than 6. This means that you couldn't possibly have a line that was 18 long in the circle.
yuppie !

Qtn is straightforward. AB becomes diameter hence subtends a 90 degree on the circumference. we get a 30-60-90 with known hypotenuse , so we get other 2 sides of triangle using ratio--hence we can calculate Area
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by El Cucu » Sun Jun 21, 2009 12:24 pm
iamcste wrote:
barnum wrote:A little bit easier to understand is that statement 2 tells you that the circ is 18pi (I assume you just left the pi out of your original post).
Can the Original poster confirm whether the circumference is 18pi or just 18

Either way, you get C but if its 18pi, perhaps problem becomes easier.
To iamcste:
Is 18 pi (have just edited the original) tks for that.

2 questions:
1) If a triangle is inscripted in a circle is a right triangle but we don´t know in this case which lenght is the diameter of the circle.

2) Can anyone explain exactly what means in data sufficiency that the figure is not drawn to scale?
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by barnum » Sun Jun 21, 2009 4:42 pm
In this question, the second statement says that the circumference is 18pi. Using the formula C = (pi)(diameter), you can solve the diameter to be 18. Since the diameter is by definition the longest line that can be drawn in a circle, the line of the triangle labeled as 18 in the picture must be the diameter.

Not drawn to scale just means that you can't look at a picture and take what you see to be necessarily an accurate measure. If a line is labeled as 18, you can trust that it is 18, but you can't just look at a line that appears to be going through the center and assume its the diameter without more information proving that it is the diameter. Or, for example, you can't look at an angle that appears to be a right triangle and assume it is a 90 degree angle without further confirmation.

Hope this helps
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