Question

In the adjoining figure ABC is a right angle triangle with ∠BAC = 90°, and AD ⊥ BC.
(i) Prove ΔADB ∼ ΔCDA.
(ii) If BD = 18 cm, CD = 8 cm find AD.
(iii) Find the ratio of the area of ΔADB is to area of ΔCDA.

Answer 1

(i) In ΔADB and ΔADC
AD = AD          ...(self)
∠ADC = 90°    ...(AD ⊥ BC)
∠ADB = 90°    ...(AD ⊥ BC)
then ∠ADC = ∠ADB
so, ΔADB ∼ ΔADC    ...(By AAA similarity)
or ΔADB ∼ ΔCDA
Hence proved.
(ii) ∵ ΔADB ∼ ΔCDA
∴ `"AD"/"BD" = "CD"/"AD"` 
`("corresponding parts of similar Δ's are proportional")`
or
AD² = BD x CD
⇒ AD² = 18 x 8  ...`(("BD" = 18),("CD" = 8),("(given)"))`
⇒ AD² = 144
AD = 12 cm
(iii) `"Area of ΔADB"/"Area of ΔCDA" = "BD"^2/"AD"^2`
`("Area theorem of similar triangles")`
= `18^2/12^2 = (18 xx 18)/(12 xx 12)`
= `(3 xx 3)/(2 xx 2) = (9)/(4)`
⇒  area (ΔADB) : area (ΔCDA) = 9 : 4.