# Thread: Ratios of areas of triangles..

1. ## Ratios of areas of triangles..

this is my problem... and i cannot even think where to begin. isnt there not enough information given??

In the figure below, AD is the perpendicular bisector of BC, AE = 9, and DE = 3.
What is the ratio of the area of triangle ABC to the area of triangle BCE?
(Hint: Area of triangle ABC = 1/2 (AD)(BC) and Area of triangle BCE = 1/2 (DE)(BC)

2. The ratio of area ABC to BCE is 4:1. That is, ABC is 4 times the area of BCE.
Yes, it can be read straight off the diagram (12:3) but here's the proof. I hope it makes sense to you. If you're still confused let me know and I'll try to simplify it more.

3. Hello, JoDan!

There's plenty of information . . .

They even spelled-out the areas of the two triangles for you.

$\displaystyle \Delta ABC \:=\:\frac{1}{2}(BC)(AD)$
. . And we know that $\displaystyle AD = 12$
Hence: .$\displaystyle \Delta ABC \:=\:\frac{1}{2}(BC)(12) \:=\:6(BC)$

$\displaystyle \Delta EBC \:=\:\frac{1}{2}(BC)(ED)$
. . And we know that $\displaystyle ED = 3$
Hence: .$\displaystyle \Delta EBC \:=\:\frac{1}{2}(BC)(3) \:=\:\frac{3}{2}(BC)$

The ratio of the areas is: .$\displaystyle \frac{\Delta ABC}{\Delta EBC} \:=\:\frac{6}{\frac{3}{2}} \:=\:4$

. . Therefore, the ratio is: .$\displaystyle 4:1$