弓形高为0.45米,跨度2.15米,长8.4米,体积是多少

Question

弓形高为0.45米,跨度2.15米,长8.4米,体积是多少

Answer 1

弓形高为H=0.45米,跨度L=2.15米,长L1=8.4米,体积V是多少? 弧半径为R,弧所对的圆心角为A,弓形面积为S. R^2=(R-H)^2+(L/2)^2 R^2=R^2-2*R*H+H^2+L^2/4 2*R*H=H^2+L^2/4 R=H/2+L^2/(8*H) =0.45/2+2.15^2/(8*0.45) =1.509米 A=2*ARC SIN((L/2)/R) =2*ARC SIN((2.15/2)/1.509) =90.86度 S=PI*R^2*A/360-R*COS(A/2)*L/2 =PI*1.509^2*90.86/360-1.509*COS(90.86/2)*2.15/2 =0.667平方米 V=S*L2=0.667*8.4=5.604立方米