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Triangle area

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by selango » Wed Jun 16, 2010 6:58 am
In triangle ABC, AD is the bisector of angle A, AB=10 cm, AC=14 cm and area of triangle ABD = 140 sq cm.
Find area of triangle ACD?

OA later
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by kvcpk » Wed Jun 16, 2010 7:05 am
Hi Selango,

Let AD be of length k. And let angle BAD=x
we know that the area of a triangle is 1/2absinC where C is the angle between a side and b side

So area of triangle ABD = 1/2 * 10*k * SinX=140
k*sinX = 140/5 = 28

now area of triangle ACD = 1/2 *k*14* sinX for similar reasons and also that ad is angular bisector.

so area of ACD = 7 * 28 = 196

OA please?
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by selango » Wed Jun 16, 2010 7:08 am
OA is indeed 196.


Is there any other approach without using trignometric function(Sin)?
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by kvcpk » Wed Jun 16, 2010 7:10 am
selango wrote:OA is indeed 196.


Is there any other approach without using trignometric function(Sin)?
Yes.. Angular bisector divides the area in the ratio of the sides.
So 10/14 = 140/x -> x = 196.

Simple way!!
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by albatross86 » Wed Jun 16, 2010 7:31 am
This problem either requires knowledge of angle bisector properties, or some very basic trigonometry. Either way I am not sure if this will ever appear in a GMAT.

NOTE: I forgot the 1/2 before the triangle areas sorry.

Here's my solution:

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