All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 47407 by ajfour last updated on 09/Nov/18
Answered by ajfour last updated on 10/Nov/18
letBD=x,∠A=ϕ,∠C=θSABCD=absinθ2+cdsinϕ2x2=a2+b2−2abcosθ=c2+d2−2cdcosϕ⇒(absinθ)dθdx=(cdsinϕ)dϕdx=2xdSdx=0⇒(abcosθ)dθdx+(cdcosϕ)dϕdx=0⇒cosθ+(cosϕ)(sinθsinϕ)=0⇒θ+ϕ=π⇒a2+b2−2abcosθ=c2+d2+2cdcosθorcosθ=a2+b2−c2−d22(ab+cd)(SABCD)max=sinθ2(ab+cd)=(ab+cd)21−[a2+b2−c2−d22(ab+cd)]2Smax=144(ab+cd)2−[(a2+b2)−(c2+d2)]2.
Commented by mr W last updated on 10/Nov/18
thankyousir!itshowstheareareachesmaximumwhenthequadrilateralisinscribedinacircle.
Terms of Service
Privacy Policy
Contact: info@tinkutara.com