Question

os lados de um triângulo retângulo estão em P.A. crescente, sabendo-se que Área mede 150. Calcule as medidas dos lados.

Answer 1

$\mathrm{Lados\ \to\ PA=\{a,b,c\}=\{a,a+r,a+2r\}}\\\\ \textbf{Pelo teorema pitag\'orico, teremos que:}\\\\ \mathrm{a^2+b^2=c^2\ \to\ a^2+(a+r)^2=(a+2r)^2}\\ \mathrm{a^2+a^2+2ar+r^2=a^2+4ar+4r^2}\\ \mathrm{2a^2-a^2+2ar-4ar+r^2-4r^2=0}\\ \mathrm{a^2-2ar-3r^2=0}\\\\ \mathrm{a=\dfrac{-(-2r)\pm\sqrt{(-2r)^2-4.1.(-3r^2)}}{2.1}=\dfrac{2r\pm\sqrt{4r^2+12r^2}}{2}}\\\\ \mathrm{a=\dfrac{2r\pm\sqrt{16r^2}}{2}=\dfrac{2r\pm4r}{2}=r\pm2r}\\\\ \mathrm{a'=r-2r=-r\ (n\~ao\ conv\'em)\ \| \ a''=r+2r=3r}$

$\textbf{Lados do tri\^angulo em fun\c{c}\~ao de r:}\\\\ \mathrm{Lados\ \to\ PA=\{3r,3r+r,3r+2r\}=\{3r,4r,5r\}}\\\\ \textbf{Calculando a raz\~ao da PA:}\\\\ \mathrm{A=150\ \to\ \dfrac{3r.4r}{2}=150\ \to\ 12r^2=300}\\\\ \mathrm{r^2=\dfrac{300}{12}\ \to\ r^2=25\ \to\ r=\sqrt{25}\ \to\ r=5}\\\\ \textbf{Portanto, os lados do tri\^angulo ser\~ao:}\\\\ \mathrm{Lados\ \to\ PA=\{3.5,4.5,5.5\}=\mathbf{\{15,20,25\}}}$