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Post subject: GMAT Geometry Posted: Sat Jul 10, 2010 3:48 am 

Joined: Sun May 30, 2010 3:15 am Posts: 424

In the figure above, the triangle ABC is inscribed in the circle with the center O and AB is a diameter of the circle. If side CB equals 2√3 and the segment OB equals 2, what is the area of triangle ABC? A. √3 B. 2√3 C. π√3 D. 2π√3 E. 8√3
(B) There are three main concepts that must be understood in order to solve this problem.
The first concept is that any triangle inscribed within a circle such that any one side of the triangle is a diameter of the circle, is a right triangle.
The second concept is that the proportions of a 306090 triangle are x, x√3, 2x.
Finally, the third concept is that the area of a triangle is equal to one half the product of its base and height: (1/2)bh.
Since the triangle is inscribed in the circle and one of its sides constitutes a diameter of the circle, the triangle must be a right triangle. Since we are told that OB, the radius of the circle, is equal to 2, the longest side of the triangle must be equal to 4. And, since its a right triangle with a hypotenuse equal to 4 and the second longest leg is equal to 2√3, it must be a 306090 triangle where the shortest side is equal to 2.
We are now ready to solve this problem. Using the third concept from above, the area must be equal to (1/2)bh. Taking the shortest side to be the base, b, and the longest of the two legs to be the height, h, we have:
(1/2)bh = (1/2)(2)(2√3) = 2√3.
So, the correct answer choice is (B). 
How do you get B?


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Gennadiy

Post subject: Re: GMAT Geometry Posted: Sat Jul 10, 2010 4:00 am 

Joined: Sun May 30, 2010 2:23 am Posts: 498

I'll rearrange reasoning for you and it might become more clear. However, if it is not, please specify what part of the reasoning is not clear for you.
Here it is: First of all we need to mention that since the triangle is inscribed in the circle and one of its sides constitutes a diameter of the circle, the triangle must be a right triangle.
OB, the radius, equals 2. So AB, hypotenuse of the triangle as also diameter of the circle, is 4. Now we know lengths of two sides of the triangle so using Pythagorean Theorem we can find the third one: AC = √(AB² – CB²) AC =√(4² – (2√3)²) = √(16 – 12) = √4 = 2
Knowing both legs of the right triangle we can find its area: (1/2) × AC × CB = (1/2) × 2 × 2√3 = 2√3
The right answer is choice (B).


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questioner

Post subject: Re: GMAT Geometry Posted: Sun Nov 28, 2010 5:01 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

The explanation calcualted the area by using (1/2)(2)(2√3). But I beleive it should be using (1/2)(2)(4) because 4 is the longest leg. Please confirm. Thanks.


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Gennadiy

Post subject: Re: GMAT Geometry Posted: Sun Nov 28, 2010 5:13 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498

The area of a triangle is 1/2 × side × height to the side.
In case of a right triangle one of legs can be considered to be a height. So the area of a right triangle can be calculated as 1/2 × leg × leg.
However the area of a right triangle can also be calculated as 1/2 × hypotenuse × height to the hypotenuse. In this case you need to calculate the height to the hypotenuse as well.


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questioner

Post subject: Re: GMAT Geometry Posted: Wed Dec 14, 2011 3:15 pm 

Joined: Sun May 30, 2010 3:15 am Posts: 424

If the longest side is 4, why do we arrange the formula using 2√3 as the longest side? If the segment OB equals 2, wouldn't we assume that the base (AB) would be 4, and therefore the area would be calculated as: (1/2)bh = (1/2)(4)(2)? I do not understand why the base is 2.
Having understood the correct response, I wonder, the proportions of the triangle, according to Pythagorean Theorem and the given data, wouldn't be 4, 4√3 and 2√3? (if we know that segment OB is 2, and the other one is 2√3?
Thanks!


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Gennadiy

Post subject: Re: GMAT Geometry Posted: Wed Dec 14, 2011 3:39 pm 

Joined: Sun May 30, 2010 2:23 am Posts: 498


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