Answer :

Given that,

ABC and BDF are two equilateral triangles


Let,


AB = BC = CA = x


Then,


BD = = DE = BF


(i) We have,


Area ( = x2


Area ( = ()2


= * x2


Area ( = Area (


(ii) It is given that triangles ABC and BED are equilateral triangles


ACB = DBE = 60o


BE AC (Since, alternate angles are equal)


Triangles BAF and BEC re on the same base BE and between the same parallels BE and AC


Therefore,


Area ( = Area (


Area ( = 2 Area ( (Therefore, ED is the median)


Area ( = Area (BAE)


(iii) Since,


are equilateral triangles


Therefore,


ABC = 60o and,


BDE = 60o


ABC = BDE


AB DE


Triangles BED and AED are on the same base ED and between the same parallels AB and DE


Therefore,


Area ( = Area (


Area ( - Area ( = Area ( - Area (


Area ( = Area (


(iv) Since,


ED is the median of


Therefore,


Area ( = 2 Area (


Area ( = 2 * Area (


Area ( Area (


Area ( = 2 Area (


(v) Let h be the height of vertex E, corresponding to the side BD on


Let H be the vertex A, corresponding to the side BC in


From part (i), we have


Area ( = Area (


* BD * h = ( * BC * H)


= (2 BD * H)


h = H (1)


From part (iii), we have


Area ( = Area (


= * PD * H


= * FD * 2h


= 2 ( * FD * h)


= 2 Area (


(vi) Area ( = Area ( + Area (


= Area ( + Area (


[Using part (iii) and AD is the median of Area


= Area ( + * 4 Area ( [Using part (i)]


Area ( = 2 Area ( (2)


Area ( = Area ( + Area (


2 Area ( + Area (


3 Area ( (3)


From above equations,


Area (= 2 Area ( + 2 * 3 Area ()


= Area ()


Hence,


Area ( = Area (


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