From the figure geometry, height of the isosceles trapezoid = a/2cos(λ) while the sum of parallel sides=[2b/sin(λ) + a/sin(λ)] and hence its area=(1/2)*[2b/sin(λ) + a/sin(λ)]*a/2cos(λ) =(a^2+2ab)/2sin(2λ) which is minimum when sin(2λ)=1 or when λ=45 deg. giving the minimum area as(a^2+2ab)/2 Ajit
From the figure geometry, height of the isosceles trapezoid = a/2cos(λ) while the sum of parallel sides=[2b/sin(λ) + a/sin(λ)] and hence its area=(1/2)*[2b/sin(λ) + a/sin(λ)]*a/2cos(λ) =(a^2+2ab)/2sin(2λ) which is minimum when sin(2λ)=1 or when λ=45 deg. giving the minimum area as(a^2+2ab)/2
ReplyDeleteAjit
"Area=(a^2+2ab)/2sin(2λ) which is minimum when sin(2λ)=1"
ReplyDeleteis so clear, you even don't need to take its derivative. Good solution!