If the perimeter of a the right-angle triangle is six times its smallest side, find the ratios of the three sides.
I tried solving it by using the normal area and volume.
$\endgroup$ 23 Answers
$\begingroup$Start by making three lengths $a$, $b$, and $\sqrt{a^2+b^2}$. Add these and set equal to $6a$ and solve for $b$ in terms of $a$. I'll leave the rest up to you.
$\endgroup$ $\begingroup$Let the sides be $a$,$b$, and $c$, with $a$ being the smallest side ad $c$ being the hypotenuse.
Using the Pythagoras theorem, $$c^2=a^2+b^2$$ The given condition is $$a+b+c=6a$$ Thus $$b+c=5a$$ $$(b+c)^2=25a^2$$ $$b^2+c^2+2bc=25a^2$$ $$a^2+2b^2+2bc=25a^2$$ $$24a^2-2b^2=2bc$$ $$12a^2-b^2=bc$$ $$(12a^2-b^2)^2=(bc)^2$$ $$144a^4+b^4-24a^2b^2=b^2(a^2+b^2)=b^2a^2+b^4$$ $$144a^4=25a^2b^2$$ Since $a\ne0$, $$144a^2=25b^2$$ That is, $$12a=5b$$ or $$\frac{a}{b}=\frac{5}{12}$$ $$c^2=a^2+b^2=(\frac{5}{12}b)^2+b^2$$ $$c^2=\frac{169}{144}b^2$$ $$c=\frac{13}{12}b$$ $$\frac{b}{c}=\frac{12}{13}$$ The ratio $a:b:c$ is thus $5:12:13$
$\endgroup$ 1 $\begingroup$Hint:
You have: $$a+b+c=6a \Rightarrow a=\frac{b+c}{5}$$ so, from $a^2+b^2=c^2$, we have: $$ (b+c)^2+25b^2=c^2 \Rightarrow 13\left(\frac{b}{c}\right)^2+\frac{b}{c}-12=0 $$ Solve this equation and you find (only) one acceptable value $k$ for $\frac{b}{c}$
substitute $b=kc$ in the perimeter and find $\frac{a}{c}$
$\endgroup$ 2$k=\frac{12}{13} \Rightarrow a:b:c=5:12:13$
