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:
Answer & explanation
Correct answer: option 2
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