Given:
Triangle ABC is similar to triangle XYZ.
In triangle ABC AB=2, BC=4, CA=3.
In triangle XYZ, XY=7, YZ=14, ZX=N.
To find:
The length of ZX.
Solution:
If two triangles are similar, then their corresponding sides are proportional.
Since [tex]\Delta ABC\sim \Delta XYZ[/tex], therefore
[tex]\dfrac{AB}{XY}=\dfrac{BC}{YZ}=\dfrac{CA}{ZX}[/tex]
[tex]\dfrac{2}{7}=\dfrac{4}{14}=\dfrac{3}{N}[/tex]
[tex]\dfrac{2}{7}=\dfrac{2}{7}=\dfrac{3}{N}[/tex]
[tex]\dfrac{2}{7}=\dfrac{3}{N}[/tex]
On cross multiplication, we get
[tex]2\times N=3\times 7[/tex]
[tex]2N=21[/tex]
Divide both sides by 2.
[tex]N=\dfrac{21}{2}[/tex]
[tex]N=10.5[/tex]
Therefore, the length of side ZX is 10.5 units.