Triangle Area Difference: Problem of the Week #80 - Oct. 7, 2013

[Jameson]

Let $ABC$ and $ABC'$ be two non congruent triangles with sides such that $AB=4$, $AC=AC'=2\sqrt{2}$ and $\angle B= 30^{\circ}$. Find the absolute value of the difference between the area of these triangles.
--------------------

 

[Jameson]

Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone

Solution (from MarkFL):

Spoiler

We can see that the absolute difference is the area of triangle $ACC'$, which we can also see is an isosceles triangle.

We may use the Law of Sines on triangle $ABC'$ to state:

\(\displaystyle \frac{\sin\left(30^{\circ} \right)}{2\sqrt{2}}=\frac{\sin(\theta)}{4}\implies\sin(\theta)=\frac{1}{\sqrt{2}}\implies \theta=\frac{\pi}{4}\)

Thus, we find:

\(\displaystyle \beta=\pi-2\theta=\frac{\pi}{2}\)

And so the area of triangle $ACC'$ is

\(\displaystyle A=\frac{1}{2}\left(2\sqrt{2} \right)^2=4\)

Hence the absolute difference in area of the two triangles is $4$ square units.