two right triangles both have equal hypotenuses and equal lengths of base (2m) are joined to make one triangle. The base of the new triangle is now 4 m (2m + 2m). what is the perimeter of the new triangle?
A math teacher solved this by assuming that the new triangle is equilateral so 30-60-90 special right triangle principle can be used. Is this assumption true?
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$\begingroup$No, they don't. Consider a triangle with side lengths $3, 4$ and $5$. This is a right triangle, but if we glue two copies of them together along the shortest edge, we will get a triangle with lengths $5, 5$ and $8$, which is obviously not equilateral.
$\endgroup$ $\begingroup$That's false. Let $\triangle ABC$ and $\triangle DEF$ be two isosceles right triangles (with respect to $B$ and $E$ respectively) such that $\triangle ABC \cong \triangle DEF$ and let $\ell>0$ be such that $|AB|=|BC|=\ell=|DE|=|EF|$, then $|AC|=|DF|=\ell \sqrt{2}$. Then the base of the "new triangle $\triangle BAF$" is $|BF|=2\ell$ but $|BA|=|AF|=\ell\sqrt{2}$. Thus $\triangle BAF$ is equilateral if and only if $$2\ell=\ell\sqrt{2} \iff \ell(2-\sqrt{2})=0 \iff \ell =0 \: \vee \: 2=\sqrt{2},$$ which is impossible.
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