Question

In figure , DEF is a right -angled triangle with ∠ E = 90 °.FE is produced to G and GH is drawn perpendicular to DE = 8 cm , DH = 8 cm ,DH = 6 cm and HF = 4 cm , find `("Ar" triangle "DEF")/("Ar" triangle "GHF")`

Answer 1

ln Δ DEF and Δ GHF,

∠ DEF = ∠ GHF  (90 ° each) 

∠ DEF = ∠ GHF    ...(common)

Δ DEF = Δ GHF     ....(AA corollary)

`therefore ("Ar" triangle "DEF")/("Ar" triangle "GHF") = "EF"^2/"HF"^2`   ......(1)

[The ration of areas of two similar triangle is equal to the ratio of square of their corresponding sides.]

In right  Δ DEF, (By Pythagoras theorem) 

DE² + EF² = DF² 

EF² = 10² - 8² 

EF² = 36 

EF = 6 

From ( 1), 

`("Ar" triangle "DEF")/("Ar" triangle "GHF") = (6/4)^2 = 9/4`

i.e. 9 : 4