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by Pharo » Sun Apr 01, 2012 1:00 am
My answer is "I and II only".

Here is my explanation:

One of the formula for the area of the triangle is (assuming a,b and c are sides and x is the angle between a and b) (a*b*sinx)/2

So setting a = 8, b = 12 we have :

(8*12*sinx)/2 = 48sinx

Now, we know that the sin function between 0-90degrees is increasing until its max output is 1 and decreasing between 90-180 and hits 0 at 180. We only take between 0-180 since no angle inside a triangle can be outside of that range.

Since the max value for the sin function is 1 (within the range); the max value for the area of our triangle is 48 and the lowest value is 0. Hence; 56 cannot be an answer but 35 and 48 can be.

Hope that helps :)
Last edited by Pharo on Sun Apr 01, 2012 1:02 am, edited 1 time in total.
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by killer1387 » Sun Apr 01, 2012 1:02 am
Maximum area possible is 0.5*8*12=48 i.e. when right angled.
hence option B
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by GMATGuruNY » Sun Apr 01, 2012 5:40 am
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by Mike@Magoosh » Tue Apr 03, 2012 11:13 am
bryan88 wrote:If two sides of a triangle are 12 and 8, which of the following could be the area of the triangle?
Dear bryan88

I see the GMAT guru posted a link, but I'll show a solution here. See my diagram below.

Imagine the side with length 12 (AB) is horizontal, and we are swinging the angle of the side with length 8 (AC) --- in the diagram, the three different C's show three different possible positions.
Of course, area = (1/2)b*h. The base of all these possible triangles is 12. When AC is perpendicular (shown as C2 in the diagram), then h = 8; otherwise, h is something less than 8.
This means the triangle with maximum area occurs when the 8 & 12 are perpendicular --- that triangle has an area of A = (1/2)b*h = (0.5)(8)(12) = 48.
Since h can be anything between 0 and 8, the area can be any value between 0 and 48. Thus, 35 is allowed as a possible value, and 48 of course is allowed --- it's the maximum possible value. The area of 56 is not possible.
Thus, answer = B

Does that make sense? Let me know if you have any further questions.

Mike :)
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