Triangle area

[selango]

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

[kvcpk]

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?

[selango]

OA is indeed 196.


Is there any other approach without using trignometric function(Sin)?

[kvcpk]

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!!

[albatross86]

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: