-
The Jock
- Master | Next Rank: 500 Posts
- Posts: 207
- Joined: Mon Oct 06, 2008 12:22 am
- Location: India
- Thanked: 5 times
- Followed by:3 members
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.
Source: Beat the GMAT math questions collectionby Papgust.
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.
Source: Beat the GMAT math questions collectionby Papgust.
