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area of a triangle question

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area of a triangle question [#permalink] New post 24 Mar 2013, 12:59
I have a really dumb question:

when calculating the area of a triangle <AOB, and AO = 2 and BO=2 and angle <AOB = 120 degrees, why can I not say the area is equal to 1/2*(2)*(2)?
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Re: area of a triangle question [#permalink] New post 24 Mar 2013, 13:04
jmuduke08 wrote:
I have a really dumb question:

when calculating the area of a triangle <AOB, and AO = 2 and BO=2 and angle <AOB = 120 degrees, why can I not say the area is equal to 1/2*(2)*(2)?


Because the formula is \frac{B*H}{2} and you don't have the height of this triangle, you can use AO or BO as base, but you still don't have the height...
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Re: area of a triangle question [#permalink] New post 25 Mar 2013, 08:10
jmuduke08 wrote:
I have a really dumb question:

when calculating the area of a triangle <AOB, and AO = 2 and BO=2 and angle <AOB = 120 degrees, why can I not say the area is equal to 1/2*(2)*(2)?



You don't have the height of the triangle.
Even if you consider either of AO or BO as the base of the triangle the other is not the height, it is just the side of the triangle as the angle is 120 degrees.

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Re: area of a triangle question [#permalink] New post 25 Mar 2013, 08:19
because 1/2*base*height is the formula...... height is calculated as the perpendicular distance from the opposite vertex to the base you are considering.... if you are doing 1/2*2*2 since the angle between two sides is not 90; observe that you are considering two sides instead of the height... So to avoid confusion remember the formula for area of triangle to be 1/2* a*b*Sin C where C is inclusive angle between a and b.....
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Re: area of a triangle question [#permalink] New post 29 Mar 2013, 09:32
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Re: area of a triangle question   [#permalink] 29 Mar 2013, 09:32
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