# The perimeter of an isosceles triangle $\triangle ABC$

An isosceles triangle $$\triangle ABC$$ is given with $$\angle ACB=30^\circ$$ and leg $$BC=16$$ $$cm$$. Find the perimeter of $$\triangle ABC$$.

We have two cases, right? When 1) $$AC=BC=16$$ and 2) $$AB=BC=16$$.

For the first case: let $$CH$$ be the altitude through $$C$$. Since the triangle is isosceles, $$CH$$ is also the angle bisector and $$\measuredangle ACH=\measuredangle BCH=15^\circ$$. How to approach the problem further? I have not studied trigonometry.

The first case.

Let $$BK$$ be an altitude of $$\Delta ABC$$.

Thus, $$BK=8,$$ $$CK=\sqrt{BC^2-BK^2}=\sqrt{16^2-8^2}=8\sqrt3$$ and $$AB=\sqrt{AK^2+BK^2}=\sqrt{(16-8\sqrt3)^2+8^2}=$$ $$=\sqrt{8^2(2-\sqrt3)^2+8^2}=8\sqrt{(2-\sqrt3)^2+1}=16\sqrt{2-\sqrt3},$$ which gives the answer: $$32+16\sqrt{2-\sqrt3}.$$

We can solve the second problem by the similar way.

• Thank you for the response! I didn't see that if we construct an altitude through $A$ or $B$ we can find it. – Katherine May 13 '20 at 17:50
• Can I ask you why $\sqrt{(16-8\sqrt3)^2+8^2}=8\sqrt{(2-\sqrt3)^2+1}$? – Katherine May 13 '20 at 17:51
• How did you see this that fast? – Katherine May 13 '20 at 17:53
• For the second case of the problem, I found $P_{\triangle ABC}=32+16\sqrt{3}$, right? – Katherine May 13 '20 at 17:59
• @LYI I added something,See now. – Michael Rozenberg May 13 '20 at 19:16

The only fact that you need is that the right triangle with sides $$1$$,$$\sqrt{3}$$ and hypothenuse $$2$$ has angles $$30$$, $$60$$.

Have this picture in mind: The right triangle $$MNP$$ with $$\angle NMP=30$$, $$\angle MPN=60$$ and $$\angle MNP=90$$ with $$NP=1$$, $$MN=\sqrt{3}$$ and $$MP=2$$.

Now, prolong the line $$\vec{NM}$$ until you reach the point $$Q$$ such that $$MQ=MP=2$$. With this construction $$QMP$$ is an isosceles triangle, so $$\angle MQP=\angle MPQ$$. Since $$\angle MQP+\angle MPQ=30$$, we have that in fact $$\angle MQP=\angle MPQ=15$$. Now, note that we have a right triangle $$QPN$$, with $$\angle NQP=15$$, and their sides are $$PN=1$$, $$QN=2+\sqrt{3}$$. By pythagoras, you have the right triangle $$PNQ$$ with $$PN=1$$, $$NQ=2+\sqrt{3}$$ and hypothenuse $$2\sqrt{2+\sqrt{3}}$$.

Going back to your first problem, you have a right triangle with angle $$15$$ and hypothenuse $$BC=16$$, By similarity of triangles you can get $$HB$$ (namely, $$\frac{HB}{BC}=\frac{1}{2\sqrt{2+\sqrt{3}}}$$) Can you get it from here?

• Even more, "the fact" that I mentioned at the beginning can be easily proved by tracing the altitude from an equilateral triangle of length side $2$. – Julian Mejia May 13 '20 at 16:50

No trigs.

Case 1. $$|AB|=|BC|=16$$.

This case is simple, $$\triangle BCE$$ is equilateral, $$|BD|,\ |CD|$$ and $$|AC|$$ can be easily found.

Case 2. $$|AC|=|BC|=16$$.

This case is just a couple of steps longer.

Extend $$BD$$ such that $$|DE|=|BD|$$. Then $$\triangle BCE$$ is equilateral.

\begin{align} \triangle BCD:\quad |BD|&=\tfrac12\,|BC|=8 ,\\ |CD|&=\sqrt{|BC|^2-|BD|^2} =8\,\sqrt3 ,\\ |AD|&=|AC|-|CD|=16-8\,\sqrt3 ,\\ |AB|&=\sqrt{|AD|^2+|BD|^2} =8\sqrt2(\sqrt3-1) . \end{align}

• Really interesting! Thank you for the response! I appreciate it. – Katherine May 13 '20 at 19:14

Can you use the cosine formula?

$$\cos(A)=\frac{b^2+c^2-a^2}{2bc}$$

• I have not studied trigonometry! – Katherine May 13 '20 at 16:28
• thats the easiest approach according to me – Dudeness May 13 '20 at 16:29
• This does not change the fact that I am not familiar with that concept. – Katherine May 13 '20 at 16:30
• Mind your language, please! I HAVE NOT STUDIED trigonometry. This includes trig functions. – Katherine May 13 '20 at 16:33
• and mind your language what? – Dudeness May 13 '20 at 16:34