求阴影部分的面积数学五星题 15
S阴影=SABC-SO-SADG
=1/2*10*20-5²π-SADG=100-25π-SADG
SADG=SAEH-SDEH-SEDG
SAEH=1/2*5*10=25
SDEH=(10*10-5²π)/4=25-25π/4
SEDG=SOEDG-SOEG
SOEDG=(nπ5²)/360
求n的话就要先就角GEO
Tan角=5/10约等于26.57
N=126.87
SOEDG=(nπ5²)/360=0.35*25π
Tan角=5/10,OE=5则
Sin角=OI/5 COS角=EI/5
EI=4.47 OI= 2.24 EG=8.94
SOEG=1/2*8.94*2.24=10.2
S阴影=SABC-SO-SADG
=1/2*10*20-5²π-SADG
=100-25π-SADG
=100-25π-(SAEH-SDEH-SEDG)
=100-25π-(25-(25-25π/4)-(SOEDG-SOEG))
=100-25π-(25π/4-(0.35*25π-10.02))
=100-25π-(6.25π-8.75π+10.02)
=89.98-22.5π
=19.29