在三棱柱ABC-A1B1C1中,AB=BC=2,角ABC=120°,侧面A1ACC1垂直底面ABC,侧棱AA1与底面AB

2个回答

  • 简要如下:

    (1)求证:BD垂直AA1

    由已知:AB=BC=2,角ABC=120°;D为AC中点,所以:BD垂直AC

    又因为:侧面A1ACC1垂直底面ABC;则:BD垂直 面A1ACC1;所以:BD垂直AA1

    (2)若角A1DC1=90°,求三棱柱ABC-A1B1C1的体积

    由已知:AB=BC=2,角ABC=120°;D为AC中点;可得:AD=√3;AC=2√3;BD=1

    过A1作A1F垂线交AC与F点,设AF=h;则:A1F=√3h;FD=(√3-h);

    利用△A1FD相似△A1DC1;可得:(A1C1)/(A1D)=(A1D)/DF;

    则有:(A1D)^2=(A1C1)*A1F=2√3*(√3-h)=6-2√3h;

    △A1FD中,(A1D)^2=(A1F)^2+(FD)^2=3h^2+(√3-h)^2=4h^2+3-2√3h;

    联立化简:3=4h^2;解得:h=√(3/4);A1F=√3h=3/2;即:三棱柱高

    所以:三棱柱ABC-A1B1C1的体积=S△ABC*A1F=1/2*2√3*1*3/2

    =3√3/2