Determina o valor de y
Decomposição de figuras - Teorema de Pitágoras: Matematicamente Falando 8 - Parte 1 Pág. 33 Ex. 12
Determina o valor de y nas seguintes figuras (as medidas indicadas estão em decímetros):
Aplicando o Teorema de Pitágoras nos sucessivos triângulos retângulos, obtém-se:
Alínea a):
\[\begin{array}{*{35}{l}}
{{2}^{2}}+{{a}^{2}}={{3}^{2}} & \Leftrightarrow & {{a}^{2}}={{3}^{2}}-{{2}^{2}} \\
{} & \Leftrightarrow & {{a}^{2}}=9-4 \\
{} & Logo, & a=\sqrt{5} \\
\end{array}\]
\[\begin{array}{*{35}{l}}
{{y}^{2}}={{8}^{2}}+{{(\sqrt{5})}^{2}} & \Leftrightarrow & {{y}^{2}}=64+5 \\
{} & \Leftrightarrow & {{y}^{2}}=69 \\
{} & Logo, & y=\sqrt{69} \\
\end{array}\]
Logo, $y=\sqrt{69}\,dm$.
Alínea b):
\[\begin{array}{*{35}{l}}
{{b}^{2}}={{1}^{2}}+{{1}^{2}} & \Leftrightarrow & {{b}^{2}}=2 \\
{} & Logo, & b=\sqrt{2} \\
\end{array}\]
\[\begin{array}{*{35}{l}}
{{y}^{2}}={{1}^{2}}+{{(\sqrt{2})}^{2}} & \Leftrightarrow & {{y}^{2}}=1+2 \\
{} & \Leftrightarrow & {{y}^{2}}=3 \\
{} & Logo, & y=\sqrt{3} \\
\end{array}\]
Logo, $y=\sqrt{3}\,dm$
Alínea c):
\[\begin{array}{*{35}{l}}
{{c}^{2}}={{1}^{2}}+{{3}^{2}} & \Leftrightarrow & {{c}^{2}}=10 \\
{} & Logo, & c=\sqrt{10} \\
\end{array}\]
\[\begin{array}{*{35}{l}}
{{d}^{2}}={{6}^{2}}+{{(\sqrt{10})}^{2}} & \Leftrightarrow & {{d}^{2}}=36+10 \\
{} & \Leftrightarrow & {{d}^{2}}=46 \\
{} & Logo, & d=\sqrt{46} \\
\end{array}\]
\[\begin{array}{*{35}{l}}
{{y}^{2}}={{3}^{2}}+{{(\sqrt{46})}^{2}} & \Leftrightarrow & {{y}^{2}}=9+46 \\
{} & \Leftrightarrow & {{y}^{2}}=55 \\
{} & Logo, & y=\sqrt{55} \\
\end{array}\]
Logo, $y=\sqrt{55}\,dm$.
