The greater triangle and the smaller ones (the two triangles inside the original one, that share the side y) are similar triangles, then we can formulate the following expressions:
• For the larger triangle and the triangle on the left
[tex]\frac{z}{9}=\frac{5}{z}[/tex]
From this equation, we can solve for z, to get:
[tex]\begin{gathered} \frac{z}{9}\times z=\frac{5}{z}\times z \\ \frac{z\times z}{9}=5\times\frac{z}{z} \\ \frac{z^2}{9}=5\times1 \\ \frac{z^2}{9}=5 \\ z^2=5\times9 \\ z^2=45 \\ z=\sqrt[]{45} \\ z=3\sqrt[]{5} \end{gathered}[/tex]
Then, z equals 3√5
• Similarly, with the larger triangle and the one on the right:
[tex]\begin{gathered} \frac{x}{9}=\frac{4}{x} \\ \end{gathered}[/tex]
From this expression, we can solve for x, like this:
[tex]\begin{gathered} \frac{x}{9}=\frac{4}{x} \\ \frac{x^2}{9}=4 \\ x^2=4\times9 \\ x^2=36 \\ x=\sqrt[]{36} \\ x=6 \end{gathered}[/tex]
Then, x equals 6
• With the triangles on the right and on the left:
[tex]\frac{y}{4}=\frac{5}{y}[/tex]
Solving for y, we get:
[tex]\begin{gathered} \frac{y}{4}=\frac{5}{y} \\ \frac{y^2}{4}=5 \\ y^2=5\times4 \\ y^2=20 \\ y=\sqrt[]{20} \\ y=2\sqrt[]{5} \end{gathered}[/tex]
Then, y equals 2√5
