Question

In figure, ABC is a triangle right angled at B and BD ⊥ AC. If AD = 4 cm, and CD = 5 cm, find BD and AB.

Answer 1

Given, ΔABC in which ∠B = 90° and BD ⊥ AC

Also, AD = 4 cm and CD = 5 cm

In ΔADB and ΔCDB,

∠ADB = ∠CDB   ...[Each equal to 90°]

And ∠BAD = ∠DBC   ...[Each equal to 90 – ∠C]

∴ ΔDBA ∼ ΔDCB   ...[By AAA similarity criterion]

Then, `("DB")/("DA") = ("DC")/("DB")`

⇒ DB² = DA × DC

⇒ DB² = 4 × 5

⇒ DB = `2sqrt(5)` cm

In right angled ΔBDC,

BC² = BD² + CD²  ...[By pythagoras theorem]

⇒ BC² = `(2sqrt(5))^2 + (5)^2`

= 20 + 25

= 45

⇒ BC = `sqrt(45) = 3sqrt(5)`

Again, ΔDBA ∼ ΔDCB,

∴ `("DB")/("DC") = ("BA")/("BC")` 

⇒ `(2sqrt(5))/5 = ("BA")/(3sqrt(5))`

∴ BA = `(2sqrt(5) xx 3sqrt(5))/5` = 6 cm

Hence, BD = `2sqrt(5)` cm and AB = 6 cm.