Question

In below Figure, ΔABC is right angled at C and DE ⊥ AB. Prove that ΔABC ~ ΔADE and Hence find the lengths of AE and DE.

Answer 1

In ΔACB, by Pythagoras theorem

AB² = AC² + BC²

⇒ AB² = (5)² + (12)²

⇒ AB² = 25 + 144 = 169

⇒ AB = `sqrt169` = 13 cm

In ΔAED and ΔACB

∠A = ∠A [Common]

∠AED = ∠ACB [Each 90°]

Then, ΔAED ~ ΔACB [By AA similarity]

`therefore"AE"/"AC"="DE"/"CB"="AD"/"AB"`          [Corresponding parts of similar Δ are proportional]

`rArr"AE"/5="DE"/12=3/13`

`rArr"AE"/5=3/13` and `"DE"/12=3/13`

`"AE"=15/13` cm and `"DE"=36/13` cm