Question Number 14905 by ajfour last updated on 05/Jun/17

Commented byajfour last updated on 05/Jun/17

Q. 14797 (construction method) given DE=a, EB=b ; find Area of △AEC. let diagonal of square=2s 2s=acos α+bcos (θ−α) s=(1/2)(acos α+bcos (θ−α)) ...(i) let ⊥ distance of E from AC be h. h=s−bcos (θ−α) h=(1/2)(acos α−bcos (θ−α)) ...(ii) tan α=((sin α)/(cos α))=((bsin θ)/(a+bcos θ)) ⇒ asin α=b(sin θcos α−cos θsin α) or asin α = bsin (θ−α) ...(iii) Area of △AEC =(1/2)(2s)h=sh =(1/4)[a^2 cos ^2 α−b^2 cos ^2 (θ−α)] [using (i) and (ii)] =(1/4)[a^2 −a^2 sin ^2 α−b^2 +b^2 sin ^2 (θ−α)] Area_(△AEC) =(1/4)(a^2 −b^2 ) . [using (iii)] .

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

god bless you mr Ajfour.it is a beautiful prove.i love it very much.thanks.

Commented byajfour last updated on 05/Jun/17

thanks sir, i was curious about the independence of the concerned area with θ .It is still not clear to me..!

Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 05/Jun/17

for what? the area computed finally whit no independence to θ.

Commented byajfour last updated on 05/Jun/17

i mean i it is lengthy, not direct or obvious as it is to mrW1..