In ΔABC, AB = AC and AL is perpendicular to BC at L. In ΔDEF, DE = DF and DM is perpendicular to EF at M. If (area of ΔABC) (area of ΔDEF) = 9 : 25, then $\frac{DM+AL}{DM-AL}$ is equal to:

4

It is given that ,

Triangle ABC and DEF are similar triangle.

So, \(\frac{Area \;of \;triangle\; ABC}{Area\; of\; triangle\; DEF}\) = \(\frac{( Side\; of \;triangle \;ABC)² }{(Side\; of\; triangle\; DEF)²}\)

\(\frac{Area \;of \;triangle\; ABC}{Area\; of\; triangle\; DEF}\) = \(\frac{( AL)² }{(DM)²}\)

\(\frac{9}{25}\) = \(\frac{( AL)² }{(DM)²}\)

\(\frac{AL}{DM}\) = \(\frac{3 }{5}\)

According to question,

\(\frac{DM+ AL}{DM - AL }\)

= \(\frac{5+3}{5-3}\) 

= 4