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Problem of the week #80 - October 7th, 2013

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Jameson

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Jan 26, 2012
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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.
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Jameson

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Jan 26, 2012
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Congratulations to the following members for their correct solutions:

1) MarkFL
2) anemone

Solution (from MarkFL):


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.
 
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