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Geometry question.....Need immediate help....

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by The Jock » Fri Oct 22, 2010 1:26 am
Question: In triangle ABC, AD is the bisector of |A, AB=10 cm, AC=14 cm and area of triangle ABD = 140 sq cm. Find area of triangle ACD.

The solution in the document is as follows:
Any angle bisector of any angle between 2 sides of a triangle divides the Area of the triangle into the ratio of sides .
Area of any triangle is 1/2 *(Product of any 2 sides of the triangle) * (Sin of Angle between those 2 sides)

Now coming to the question at concern.
Here area of ABD => 140 = 1/2*(AB * AD) *(Sin of angle BAD) ---eqn (1)
Area of ACD = 1/2*(AC*AD) * (Sin of angle DAC) ---eqn(2)
angle DAC = angle BAD ---eqn(3) as angle A is bisected

Using eqn 1 and 2 and 3, gives 196 as area of ACD.

But I am not able to get it fully and want another explanation.
Thanks in advance.

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by Rahul@gurome » Fri Oct 22, 2010 2:06 am
Area of triangle = ½ * product of any 2 sides * sine of angle between these two sides.
Let angle BAD be x.
So angle DAC is also x.
So area of triangle ABD is ½ * AB * AD * sin x = 140.
Or ½ * 10 * AD * sin x = 140.
So AD*sin x = 28.
Now area of triangle ACD is ½ * AC * AD * sin x = ½ * 14 * AD * sin x = 7 * 28 = 196.
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by N:Dure » Sat Oct 23, 2010 7:02 am
lol!
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by goyalsau » Sat Oct 23, 2010 9:01 am
Rahul@gurome wrote:
Area of triangle = ½ * product of any 2 sides * sine of angle between these two sides.

.
Awesome, Awesome....................
It always been learning at BTG,
I never knew about this formula
Area of triangle =
1/2 * product of any two sides * sine of angle between them

But i was thinking it should have been product of two adjacent sides and that must share a common vertices as per this question .

Rahul please correct me if am wrong.
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by Rahul@gurome » Sat Oct 23, 2010 11:06 am
goyalsau wrote: ...But i was thinking it should have been product of two adjacent sides and that must share a common vertices as per this question .

Rahul please correct me if am wrong.
Well, any two sides of a triangle always share a common vertex. Isn't it?
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by Ian Stewart » Sat Oct 23, 2010 12:37 pm
You will never need to use sines or cosines on a real GMAT question. Just ignore this one and move on to more relevant material.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

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