Question

Q. 13
From the Triangles chapter. Please help!

Answer 1

Given :- ∆ ABC is an equilateral triangle . AD _|_ BC . With AD as base ∆ AED is constructed .

To prove :- ar (∆ADE) : ar(∆ABC) = 3 : 4

Proof :----- Since ∆ ABC is equilateral

Therefore ----------

AB = BC = AC

Since AD is the altitude of ∆ ABC


:. AD = √3/4 AB .......................................... 1


Since ∆ AED is also an equilateral triangle .

AD = AE = ED


Therefore ---------------

AD = AE = ED = √3/4 AB .................................. ( from 1 )



In ∆ BAC

/_ BAC = 60°

Since AD is the bisector


Therefore --------

/_ BAD = /_ CAD = 30° ........................... 2


Since ∆ AED is equilateral

Therefore --------------

/_ DAE = 60° ....................... 3


FROM 2 & 3

/_ BAD = /_ EAC = 30° ................................ 4


AND ALSO ------

/_ ACB = /_ AED = 60° .................. ( since ∆ ABC & ∆ AED are equilateral ) .............5


In ∆ ABC & ∆ AED

/_ BAD = /_ EAC ................... ( from 4 )

/_ ACB = /_ AED .................... ( from 5 )


So by A.A test

∆ ABC is similar to ∆ ADE .



So ar (∆ ADE ) : ar (∆ ABC ) = ( AD / AB )²


= ( √3/4 AB / AB )²


= 3/4 AB² / AB²

= 3/4. OR. 3:4




HENCE PROVED ............


HOPE IT IS MOST HELPFUL!!!!!!!!!!