Same proof as number 250: Say base of area B is x. Say area D is a multiple of area B by factor k. Then base area D = k*x and A(D) = k^2*A(B). The base of area C is then x + kx = x*(k+1). So area C is a multiple of B by factor k+1 and A(C)=(k+1)^2*A(B). The shaded area = A(C)-A(B)-A(D) = ((k+1)^2-1-k^2)*A(B) = 2k*A(B). Area E is the result of Area(B) first horizontally stretched by factor k and then vertically multiplied by factor 2, so A(E) = A(B)*k*2 = 2k*A(B). QED. (Sorry for my poor English... I'm Dutch...)
http://geometri-problemleri.blogspot.com/2010/12/problem-96-ve-cozumu.html
ReplyDeleteSame proof as number 250:
ReplyDeleteSay base of area B is x.
Say area D is a multiple of area B by factor k.
Then base area D = k*x and A(D) = k^2*A(B).
The base of area C is then x + kx = x*(k+1).
So area C is a multiple of B by factor k+1 and A(C)=(k+1)^2*A(B).
The shaded area = A(C)-A(B)-A(D) = ((k+1)^2-1-k^2)*A(B) = 2k*A(B).
Area E is the result of Area(B) first horizontally stretched by factor k and then vertically multiplied by factor 2, so A(E) = A(B)*k*2 = 2k*A(B).
QED.
(Sorry for my poor English... I'm Dutch...)
Your proof and English are just great!
ReplyDeleteThank you Henkie.