x = 37.5 (or) [tex]\frac{75}{2}[/tex]
Solution:
Given [tex]\triangle A B C \sim \triangle D B E[/tex]
AC = 50, DE = 30, EC = 25, BE = x, BC = 25 + x
To find the value of x:
Property of similar triangles:
If two triangles are similar then the corresponding angles are congruent and the corresponding sides are in proportion.
[tex]$\frac{BE}{BC} =\frac{DE}{AC}[/tex]
[tex]$\frac{x}{25+x} =\frac{30}{50}[/tex]
Do cross multiplication, we get
[tex]50x=30(25+x)[/tex]
[tex]50x=750+30x[/tex]
Subtract 30x from both sides of the equation.
[tex]20x=750[/tex]
Divide by 20 on both sides of the equation, we get
x = 37.5 (or) [tex]\frac{75}{2}[/tex]
Hence the value of x is 37.5 or [tex]\frac{75}{2}[/tex].
