Given:
ΔABC [tex]\sim[/tex] ΔDEF
To find:
The length of median CP
Solution:
In ΔABC,
AP = 12, BP = 12 and PC = 3x - 12
In ΔDEF,
DQ = 16, QE = 16 and FQ = 2x + 8
If two triangles are similar, then their median is proportional to the corresponding sides.
[tex]$\Rightarrow \frac{AP}{DQ} =\frac{PC}{FQ}[/tex]
[tex]$\Rightarrow \frac{12}{16} =\frac{3x-12}{2x+8}[/tex]
Do cross multiplication.
[tex]$\Rightarrow 12(2x+8)=16(3x-12)[/tex]
[tex]$\Rightarrow 24x+96=48x-192[/tex]
Add 192 on both sides.
[tex]$\Rightarrow 24x+96+192=48x-192+192[/tex]
[tex]$\Rightarrow 24x+288=48x[/tex]
Subtract 24x from both sides.
[tex]$\Rightarrow 24x+288- 24x=48x- 24x[/tex]
[tex]$\Rightarrow 288=24x[/tex]
Divide by 24 on both sides.
⇒ 12 = x
Substitute x = 12 in CP.
CP = 3(12) - 12
= 36 - 12
= 24
The length of median CP is 24.
