Given:
ΔABC undergoes a dilation by a scale factor and comes as ΔA'B'C'.
To show that both the triangles are similar.
Formula
- By the condition of similarity we get,
If two triangles have three pairs of sides in the same ratio, then the triangles are similar.
- By Pythagoras theorem we get,
[tex]Hypotenuse^2 = Base^2+Height^2[/tex]
Now,
In ΔABC,
AB = 18 unit
BC = 10 unit
So, [tex]AC^2 = AB^2+BC^2[/tex]
or, [tex]AC^2 = 18^2+10^2[/tex]
or, [tex]AC = \sqrt{424}[/tex]
Again,
In ΔA'B'C'
A'B' = 9 unit
B'C' = 5 unit
So, [tex]A'C' ^2 = A'B'^2+B'C'^2[/tex]
or, [tex]A'C'^2 = 9^2+5^2[/tex]
or, [tex]A'C' = \sqrt{106}[/tex]
Now,
[tex]\frac{AB}{A'B'} = \frac{18}{9} = 2[/tex]
[tex]\frac{BC}{B'C'} = \frac{10}{5} = 2[/tex]
[tex]\frac{AC}{A'C'} = \frac{\sqrt{424} }{\sqrt{106} } = 2[/tex]
Hence,
All the ratios are equal.
Therefore, we can conclude that,
ΔABC and ΔA'B'C' are similar.
