What is x given triangleABC ~triangleDBE?
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x = 37.5 (or) [tex]\frac{75}{2}[/tex]
Solution:
Given [tex]\triangle A B C \sim \triangle D B E[/tex].
Let us take BE = x and BC = 25 + x.
To determine the value of x:
If two triangles are similar then the corresponding angles are congruent and the corresponding sides are in proportion.
[tex]$\frac{AC}{DE}=\frac{B C}{B E}[/tex]
[tex]$\frac{50}{30} =\frac{25+x}{x}[/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]20 x=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].