Geometry question, please help?

enter image source here

↳Redirected from "What is the freezing point depression?"
1 Answer
CW
Sep 24, 2017

#=R^2/12(pi-3sqrt3+6)#

Explanation:

enter image source here
As #OA=OB=R, and AB=Rsqrt2#,
#=> angleAOB=90^@#
Use Law of cosines to find #angleAOC#,
#cosangleAOC=(R^2+R^2-(sqrt3R)^2)/(2R^2)#
#=> cosangleAOC=-(R^2)/(2R^2)=-1/2#
#=> angleAOC=cos^-1(-1/2)=120^@#
Yellow area = area of sector #OAC -# green area #-# area #DeltaOAC#
Area of sector #OAC = piR^2(120)/360=color(red)((piR^2)/3)#
Green area #=#area of sector #OAB - # area #DeltaOAB#
#= color(red)((piR^2)/4-R^2/2)#
Area #DeltaOAC=1/2R^2sin120=color(red)((sqrt3R^2)/4)#

Hence, yellow area
#=(piR^2)/3-(R^2sqrt3)/4-(piR^2)/4+R^2/2#
#=R^2(pi/3-sqrt3/4-pi/4+1/2)#
#=R^2(pi/12-sqrt3/4+1/2)#
#=R^2/12(pi-3sqrt3+6)#

Related questions
Impact of this question
1465 views around the world
You can reuse this answer
Creative Commons License