In Fig.. (a) ΔABC is right angled at C and DE ⊥ AB. Prove that �

1.	In Fig.. (a) ΔABC is right angled at C and DE ⊥ AB. Prove that ΔABC ~ ΔADE and hence find the lengths of AE and DE.
Answer» In ΔABC, by Pythagoras theorem AB² = AC² + BC² Or, AB² = 5² + 12² Or, AB² = 25 + 144 Or, AB² = 169 Or AB = 13 (Square root both side) In Δ AED and Δ ACB ∠A = ∠A (Common) ∠AED = ∠ACB (Each 90°) Then, Δ AED ~ Δ ACB(Corresponding parts of similar triangle are proportional) So, AE/AC = DE/ CB =AD/ AB Or, AE/5 = DE/12 = 3/13 Or, AE/5 = 3/13 and DE/12 = 3/13 Or, AE = 15/13cm and DE = 36/13 cm

In Fig.. (a) ΔABC is right angled at C and DE ⊥ AB. Prove that ΔABC ~ ΔADE and hence find the lengths of AE and DE.

Answer»

In ΔABC, by Pythagoras theorem

AB² = AC² + BC²

Or, AB² = 5² + 12²

Or, AB² = 25 + 144

Or, AB² = 169

Or AB = 13 (Square root both side)

In Δ AED and Δ ACB

∠A = ∠A (Common)

∠AED = ∠ACB (Each 90°)

Then, Δ AED ~ Δ ACB(Corresponding parts of similar triangle are proportional)

So, AE/AC = DE/ CB =AD/ AB

Or, AE/5 = DE/12 = 3/13

Or, AE/5 = 3/13 and DE/12 = 3/13

Or, AE = 15/13cm and DE = 36/13 cm