we are given that all the triangles are congruent, which means that corresponding sides are at the same ratio with each other.
First, we will use the Pythagorean theorem to determine segment EC:
[tex]\begin{gathered} EC=\sqrt[]{(25)^{2^{}}+(15)^2} \\ EC=\sqrt[]{400} \\ EC=20 \end{gathered}[/tex]Since Triangles EBC and EXC are congruent we have:
[tex]\frac{EX}{EC}=\frac{EB}{BC}[/tex]Now we solve for EX by multiplying both sides by EC:
[tex]EX=EC\times\frac{EB}{BC}[/tex]Now we replace the known values:
[tex]\begin{gathered} EX=(20)(\frac{15}{25}) \\ EX=12 \end{gathered}[/tex]Therefore, b.) EX = 12.
Now we use the Pythagorean theorem in triangle EXC.
[tex]XC=\sqrt[]{(EC)^2-(EX)^2}[/tex]Replacing the known values:
[tex]XC=\sqrt[]{(20)^2-(12)^2}[/tex]Solving the operations:
[tex]\begin{gathered} XC=\sqrt[]{400-144} \\ XC=\sqrt[]{256}=16 \end{gathered}[/tex]we have that:
[tex]BX=BC-XC[/tex]Replacing the known values:
[tex]\begin{gathered} BX=25-16 \\ BX=9 \end{gathered}[/tex]Therefore, a.) BX = 9.
We have the following relationship:
[tex]XC=XY+YZ+ZC[/tex]we also have:
[tex]\begin{gathered} \frac{YZ}{XY}=\frac{1}{3} \\ \frac{YZ}{ZC}=\frac{1}{4} \end{gathered}[/tex]Rewritten we get:
[tex]\begin{gathered} XY=3YZ \\ ZC=4YZ \end{gathered}[/tex]Replacing in the relationship we get:
[tex]\begin{gathered} XC=3YZ+YZ+4YZ \\ XC=8YZ \end{gathered}[/tex]Solving for YZ:
[tex]\frac{XC}{8}=YZ[/tex]Replacing the known value:
[tex]\begin{gathered} \frac{16}{8}=YZ \\ 2=YZ \end{gathered}[/tex]Therefore, h.) YZ = 2
Using the proportions we get:
[tex]XY=3(2)=6[/tex]Therefore, e.) XY = 6
[tex]ZC=4(2)=8[/tex]Therefore, k.) ZC = 8
Now we use the following proportion:
[tex]\frac{GC}{ZC}=\frac{EC}{XC}[/tex]Solving for GC:
[tex]GC=ZC\times\frac{EC}{XC}[/tex]Replacing the known values:
[tex]GC=(8)(\frac{20}{16})[/tex]Solving the operations:
[tex]GC=10[/tex]Therefore, i.) GC = 10.
We have the following proportion:
[tex]\frac{FG}{GC}=\frac{YZ}{ZC}[/tex]Solving for FG:
[tex]FG=GC\times\frac{YZ}{ZC}[/tex]Replacing the known values:
[tex]FG=(10)(\frac{1}{4})=2.5[/tex]Therefore, f.) FG = 2.5
We also have
[tex]\frac{XY}{YZ}=\frac{EF}{FG}[/tex]Solving for EF:
[tex]EF=FG\times\frac{XY}{YZ}[/tex]Replacing the values:
[tex]EF=(2.5)(\frac{3}{1})=7.5[/tex]We have the relationship:
[tex]\frac{EC}{EX}=\frac{FC}{FY}[/tex]Solving for FY:
[tex]FY=FC\times\frac{EX}{EC}[/tex][tex]FY=(FG+GC)\times\frac{EX}{EC}[/tex]Replacing:
[tex]FY=(2.5+10)\times\frac{12}{20}[/tex]Solving the operations:
[tex]FY=7.5[/tex]Therefore, d.) FY=7.5.
Finally, we can use the following proportion:
[tex]\frac{GC}{GZ}=\frac{FC}{FY}[/tex]Solving for GZ:
[tex]GZ=GC\times\frac{FY}{FC}[/tex]Replacing we get:
[tex]GZ=10\times(\frac{7.5}{12.5})[/tex]Solving the operations:
[tex]GZ=6[/tex]Therefore, g.) GZ = 6.
