"Find the missing side lengths in each pair of similar figures." Someone explain please

1. For triangle ABC, the value of 'x' and 'y' are 15 and 20 respectively.
2. For triangle XYZ, the value of 'x' and 'y' are 17 and 8 respectively.
A triangle is a two-dimensional geometrical figure that has three sides, three interior angles, and three vertices.
1. ΔABC and ΔXYZ are similar.
Therefore, [tex]\frac{AB}{XY} = \frac{25}{10}[/tex]
⇒ [tex]\frac{AB}{XY} = 2.5[/tex]
Here, [tex]AC = x, XZ = 6[/tex]
Therefore, [tex]\frac{AC}{XZ} = 2.5[/tex]
⇒ [tex]AC = 2.5(XZ)[/tex]
⇒ [tex]x = (2.5)(6)[/tex]
⇒ [tex]x = 15[/tex]
Again, [tex]CB = y, ZY = 8[/tex]
[tex]\frac{CB}{ZY}= 2.5[/tex]
⇒ [tex]\frac{y}{8} = 2.5[/tex]
⇒ [tex]y = (2.5) (8)[/tex]
⇒ [tex]y = 20[/tex]
2. For ΔABC
BC
[tex]= \sqrt{AB^{2} - AC^{2}}\\= \sqrt{34^{2} - 16^{2}}\\= \sqrt{900}\\= 30[/tex](Side can't be negative.)
Here, ΔABC and ΔXYZ are similar.
Therefore,
[tex]\frac{BC}{ZY}\\= \frac{30}{15}\\= 2[/tex]
Now, [tex]AB = 34, XY = x[/tex]
Therefore, [tex]\frac{AB}{XY} = 2[/tex]
⇒ [tex]\frac{34}{x} = 2[/tex]
⇒ [tex]x = \frac{34}{2}[/tex]
⇒ [tex]x = 17[/tex]
Again, [tex]AC = 16, XZ = y[/tex]
Therefore, [tex]\frac{AC}{XZ} = \frac{16}{y} = 2[/tex]
⇒ [tex]\frac{16}{y} = 2[/tex]
⇒ [tex]y = \frac{16}{2}[/tex]
⇒ [tex]y = 8[/tex]
Learn more about a triangle here: https://brainly.com/question/2338119
#SPJ3