Стереометрия

[Milenaa]

дали може още помощ не съм нахална вече

[Nathi123]

Нека [tex]\angle ACB=\angle A_{1 }C_{1 }B_{1 }=90^\circ ; \angle ABC=30^\circ \Rightarrow AB=c.[/tex]
Построяваме [tex]CH\bot AB\Rightarrow C_{1 }H \bot AB[/tex] Т-ма за 3-те перпенд. [tex]\Rightarrow \angle CHC_{1 }=45^\circ\Rightarrow \angle CC_{1 }H=45^\circ[/tex]
[tex]\Rightarrow CC_{1 }=CH ; \Delta ABC\Rightarrow AC=\frac{1}{2}AB=\frac{c}{2} ; BC=\frac{c\sqrt{3}}{2}\Rightarrow 2S_{\Delta ABC}=\frac{c^{2}\sqrt{3}}{4}=CH.c\Rightarrow CH=CC_{1 }=\frac{c\sqrt{3}}{4}[/tex].
[tex]V_{ABCC_{1 }}=\frac{S_{\Delta ABC }.CH}{3}=\frac{c^{2}\sqrt{3}.c\sqrt{3}}{24.4}=\frac{c^{3}}{32}.[/tex]

[amsara]

35.
[tex]BF=CH=x=(21-11):2=5 cm; BF \bot AD; CH \bot AD[/tex]
[tex]\Delta HDC: 5^2+x^2=13^2 <=>x^2=144; x=12 cm[/tex]
[tex]B_{ABCD}=\frac{11+21}{2}.12=192 cm^2[/tex]
[tex]AC=x; AH^2+12^2=AC^2 <=> 16^2+12^2=AC^2; AC=20cm[/tex]
[tex]180=20.CC_{1}; CC_{1}=9cm=h[/tex]
[tex]S_{1}=S+2B[/tex]
[tex]S=P.h=58.9=522 sm^2[/tex]
[tex]S_{1}=522+2.192=906 cm^2[/tex]

[amsara]

36.
[tex]a=9x; b=10x, c=17x[/tex]
[tex]p=36x/2=18x[/tex]
[tex]B=\sqrt{18x.(18x-9x)(18-10x)(18-17x)}=36x^2[/tex]
[tex]36x^2=4; x^2=\frac{1}{9}; x=\frac{1}{3} m[/tex]
[tex]a=9.\frac{1}{3}=3m; a.h=9 => h=3m[/tex]
[tex]V=B.h=4.3=12m^3[/tex]

[Milenaa]

БЛАГОДАРЯ ВИ!