Step-by-step explanation:
(c)
[tex] In\: \triangle PLM \: \&\: \triangle PLM\\
\angle PLM \cong\triangle PFG.. (corresponding \:\angle s) \\
\angle LPM \cong \angle FPG.. (common \: \angle s) \\
\therefore \triangle PLM\sim\triangle PFG.. (by\: AA\:Postulate) \\
PF = PL + LF = 2 + 3 = 5[/tex]
By area of similar triangle theorem:
[tex] \frac {A(\triangle PLM)}{A(\triangle PFG)}= \frac{PL^2}{PF^2}\\
\therefore \frac {A(\triangle PLM)}{A(\triangle PFG)}= \frac{2^2}{5^2}\\
\therefore \frac {A(\triangle PLM)}{A(\triangle PFG)}= \frac{4}{25}\\
A(\triangle PLM)\: :\: A(\triangle PFG) = 4 \: :\: 25
[/tex]
