Area of a triangle

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January 8th 2011, 11:46 PM
griffen

Area of a triangle

Hello. I am stuck on this problem :(

We are given triangle ABC and point E on segment AC, such that <ABE = 75 deg and <EBC= 75 deg. Also, AB = 4 and BC = 6. Find the areas of triangles ABE and EBC.

Now at first I thought that the angle bisector formed a right angle but now I think it doesn't. By the angle bisector theorem, we can say that $\displaystyle \frac{4}{AE}=\frac{6}{EC}$ but that doesn't get me any further. I can find the area of the triangle by using the law of cosines but that also doesn't help me. I feel like the fact that there is an angle bisector should be a big part of the solution, but I'm stumped.

Any hints?
January 9th 2011, 12:31 AM
alexmahone

Quote:

Originally Posted by griffen

Hello. I am stuck on this problem :(

We are given triangle ABC and point E on segment AC, such that <ABE = 75 deg and <EBC= 75 deg. Also, AB = 4 and BC = 6. Find the areas of triangles ABE and EBC.

Now at first I thought that the angle bisector formed a right angle but now I think it doesn't. By the angle bisector theorem, we can say that $\displaystyle \frac{4}{AE}=\frac{6}{EC}$ but that doesn't get me any further. I can find the area of the triangle by using the law of cosines but that also doesn't help me. I feel like the fact that there is an angle bisector should be a big part of the solution, but I'm stumped.

Any hints?

$area\Delta ABC=\frac{1}{2}*AB*BC*sin C=\frac{1}{2}*4*6*sin 150^o$

$\frac{AE}{EC}=\frac{4}{6}\implies\frac{area\Delta ABE}{area\Delta EBC}=\frac{4}{6}$

Can you finish off?
January 9th 2011, 01:02 AM
griffen

Thanks!

I got before that the area could be calculated that way, I just didn't know how to use it. I get the area of ABC to be 6. Now if the legs of the triangles are in that ratio and have the same height, then the areas also are in the same ratio. That was the insight I was missing!

$2x+3x=6$

The ratio reduces to 3:2, so 3x+4x=6 for some x since the two smaller triangles add up to the bigger one. This yields $x=\frac{6}{5}$ and this means that the two triangles have areas $\frac{12}{5},\frac{18}{5}$

Look good?
January 9th 2011, 06:59 AM
griffen

That isn't right I think now. I'm stuck again. I don't see how to relate the fact that the area of ABE+EBC=ABC and that AE/EC=4/6. I've been thinking all day. Can I have a hint please? I've searched online for a while too and can't see anything that's clicking.
January 9th 2011, 07:04 AM
alexmahone

Quote:

Originally Posted by griffen

That isn't right I think now. I'm stuck again. I don't see how to relate the fact that the area of ABE+EBC=ABC and that AE/EC=4/6. I've been thinking all day. Can I have a hint please? I've searched online for a while too and can't see anything that's clicking.

Why are you stuck? Your answer in post #3 is correct.
January 9th 2011, 08:11 AM
griffen

Quote:

Originally Posted by alexmahone

Why are you stuck? Your answer in post #3 is correct.

Oh wow, haha. I guess I just don't see how $\displaystyle \frac{AE}{EC}=\frac{4}{6}\implies\frac{area\Delta ABE}{area\Delta EBC}=\frac{4}{6}$
January 9th 2011, 08:17 AM
alexmahone

Quote:

Originally Posted by griffen

Oh wow, haha. I guess I just don't see how $\displaystyle \frac{AE}{EC}=\frac{4}{6}\implies\frac{area\Delta ABE}{area\Delta EBC}=\frac{4}{6}$

Draw a diagram. Note that triangles ABE and EBC have the same altitude BX (say).

$\displaystyle \frac{area\Delta ABE}{area\Delta EBC}=\frac{\frac{1}{2}*AE*BX}{\frac{1}{2}*EC*{BX}} =\frac{AE}{EC}$