Question

Triangle ABC is an isosceles triangle in which AB = AC = 13 cm and BC = 10 cm. AD is
perpendicular to BC. If CE = 8 cm and EF ⊥ AB, find:

i)`"area of ADC"/"area of FEB"`       ii)`"area of ΔAFEB"/"area of ΔABC"`

Answer 1

(i) AB = AC(Given)
∴ ∠FBE = ∠ACD
∠BFE = ∠ADC
∆EFB ~ ∆ADC     (AA similarity)

`∴(ar(Δ"ADC"))/(ar(Δ"EFB"))=(("AC")/("BE"))^2`

`=(("AC")/("BC"+"CE"))^2`

`(13/18)^2=169/324`                ................................(1)   

ii) Similarly, it can be proved that ∆ADB ~ ∆EFB

`∴(ar(Δ"ADB"))/(ar(Δ"EFG")`= `(("AB")/("BE"))^2`

`=(13/18)^2`

`=169/324`       .......................... (2) 

From (1) and (2),

`(ar(Δ"ABC"))/(ar(Δ"EFB"))=(169+169)/324=338/324=169/162`

∴ar(∆EFB) : ar(∆ABC) = 162 : 169