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In the triangle above, we have $AC = 12, \ \ BD = 10, \ \ CD = 6$.

Find the area of $\triangle ABD$.


This was part of an exam where you are not allowed to use a calculator or any other digital tools.

My progress

Given $AC, CD, \measuredangle C$, I was able to find the length of $AD$ using the law of cosines. It yielded $AD = \sqrt{180} = 6\sqrt5$.

From here, I decided to try and use the law of sines to find $\measuredangle A$ as it would be the same in $\triangle ABD$ and $\triangle ACD$.

I got $$\frac{\sin A}{CD} = \frac{\sin C}{AD}$$.

This yields $$\sin A = \frac{CD \sin C}{AD} = \frac{6}{\sqrt{180}}$$.

Then I need to calculate $\arcsin\frac{6}{\sqrt{180}}$, which I can't, because I can't use a calculator for this problem. (Well, technically I can, but I want to solve this within the constraints placed on the participants.)

From here, I am stuck. In $\triangle ABD$, I know now two of the lengths, but I don't know any of the angles.


Am I overlooking an easy inverse sine here? Or is there another way to calculate this area?

Thanks in advance for any help!

  • $\begingroup$ If this is part of an exam, why are you asking for help? $\endgroup$ – N. F. Taussig Mar 21 '17 at 20:07
  • $\begingroup$ @N.F.Taussig - It was an exam given in August of 2016. I'm preparing for the 2017 one ;) $\endgroup$ – Alec Mar 21 '17 at 20:08

Find $BC$ in the right triangle $BCD$; you have the hypotenuse is $10$ and the other side is $6$ so $BC=8$.

Area $\triangle ACD = \frac12\cdot 12\cdot 6 = 36$.

Area $\triangle BCD = \frac12\cdot 8\cdot 6 = 24$.

Area $\triangle ABD = 36-24 = 12$.

  • $\begingroup$ Damn... I really blinded myself on this one. Thanks! $\endgroup$ – Alec Mar 21 '17 at 20:11

Calculate $BC$ with the pythagorean theorem, and subtract the areas of the two right angled triangles.


$$BC=\sqrt{10^2-6^2}=8$$ Hence, $$AB=12-8=4$$ and $$S_{\Delta ABD}=\frac{4\cdot6}{2}=12$$


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