In ΔABC and ΔDEF, it is being given that: AB = 5 cm, BC = 4 cm and C

1.	In ΔABC and ΔDEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL ⊥BC and DM ⊥ EF, find AL : DM.
Answer» Since \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\) = \(\frac{1}{2}\) Then, ΔABC ~ ΔDEF (By SS similarity) Now, In ΔABL ~ ΔDEM ∠B = ∠E (ΔABC ~ΔDEF) ∠ALB = ∠DME (Each 90°) Then, ΔABL ~ ΔDEM (By SS similarity) So, \(\frac{AB}{DE}\) = \(\frac{AL}{DM}\) (Corresponding parts of similar triangle area proportion) or \(\frac{5}{10}\) = \(\frac{AL}{DM}\) or, \(\frac{1}{2}\) = \(\frac{AL}{DM}\)

In ΔABC and ΔDEF, it is being given that: AB = 5 cm, BC = 4 cm and CA = 4.2 cm; DE = 10 cm, EF = 8 cm and FD = 8.4 cm. If AL ⊥BC and DM ⊥ EF, find AL : DM.

Answer»

Since \(\frac{AB}{DE}\) = \(\frac{BC}{EF}\) = \(\frac{AC}{DF}\) = \(\frac{1}{2}\)

Then, ΔABC ~ ΔDEF (By SS similarity)

Now, In ΔABL ~ ΔDEM

∠B = ∠E (ΔABC ~ΔDEF)

∠ALB = ∠DME (Each 90°)

Then, ΔABL ~ ΔDEM (By SS similarity)

So, \(\frac{AB}{DE}\) = \(\frac{AL}{DM}\) (Corresponding parts of similar triangle area proportion)

or \(\frac{5}{10}\) = \(\frac{AL}{DM}\)

or, \(\frac{1}{2}\) = \(\frac{AL}{DM}\)