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[Гост]

Здравейте, как се решават този набор от задачи? Не ми се получават отговорите, дадени в помагалото. Да не би да има грешка?

[ammornil]

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[tex]\boxed{6}\quad S=24[cm^{2}], S=\frac{2}{3}\cdot{}S_{1} \\ S_{1}=\frac{3}{2}\cdot{}S, \quad S_{1}=S+2B \\ B=\frac{S_{1}-S}{2}=\frac{\frac{3}{2}\cdot{}S-S}{2}=\frac{\frac{1}{2}\cdot{}S}{2}=\frac{S}{4}=6[cm^{2}][/tex]

[tex]\boxed{7}\quad p(ABC)\| p(A_{1}B_{1}C_{1}),\quad AA_{1}\|BB_{1}\|CC_{1},\quad AA_{1}\bot{p(ABC)} \quad \triangle{ABC}\cong\triangle{A_{1}B_{1}C_{1}} \\S_{1,призма}=S_{1,(ABCA_{1}B_{1}C_{1})}=340[dm^{2}],\\ \triangle{ABC}, AB=12[dm], BC=9[dm] \\ \text{(1случай)} \quad AB\bot{BC} \Rightarrow AC=\sqrt{AB^{2}+BC^{2}}=15[dm] \\ h_{AC}=7,2[dm], S_{ABC}=\frac{AB\cdot{BC}}{2}=\frac{AC\cdot{h_{AC}}}{2} \Leftrightarrow AB\cdot{BC}=AC\cdot{h_{AC}} \rightarrow \text{ вярно равенство } \Rightarrow \exists{\triangle{АBC}} \\ S_{ABC}=54[dm^{2}]\\ S_{1,призма}=2\cdot{S_{ABC}}+P_{ABC}\cdot{h_{призма}} \Rightarrow h_{призма}=\frac{S_{1,призма}-2\cdot{S_{ABC}}}{AB+BC+AC}=\frac{340-108}{12+9+15}=\frac{232}{36}=6\frac{4}{9}[dm]\\ \phantom{q} \\ \text{(2случай)} \quad AB\not\perp{BC} \Rightarrow AC=\sqrt{AB^{2}-BC^{2}}=3\sqrt{7}[dm] \\ h_{AB}=7,2[dm], \rightarrow AC\cdot{BC}\ne AB\cdot{h_{AB}}, \quad S_{ABC}=\frac{AC\cdot{BC}}{2}=\frac{AB\cdot{h_{AB}}}{2} \rightarrow \text{ не е вярно равенство } \Rightarrow \nexists{\triangle{АBC}}[/tex]

[tex]\boxed{8}\quad B=S_{1}-S=60[cm^{2}], H=32[cm] \Rightarrow V=\frac{1}{3}\cdot{}B\cdot{}H=\frac{1}{3}\cdot{60}\cdot{32}=640[cm^{3}][/tex]

[tex]\boxed{9}\quad V=25[cm^{3}], H=3[cm], k=3,9[cm] \\ V=\frac{1}{3}\cdot{B}\cdot{H} \Rightarrow B=\frac{3\cdot{V}}{H}=25[cm^{2}] \\ B=a^{2} \Rightarrow a=5[cm] \text{ основен ръб на правилната четириъгълна пирамида} \\ S=4\cdot{a}\cdot{k}=78[cm^{2}] \\ S_{1}=S+B=78+25=103[cm^{2}][/tex]

[Гост]

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[tex]\boxed{6}\quad S=24[cm^{2}], S=\frac{2}{3}\cdot{}S_{1} \\ S_{1}=\frac{3}{2}\cdot{}S, \quad S_{1}=S+2B \\ B=\frac{S_{1}-S}{2}=\frac{\frac{3}{2}\cdot{}S-S}{2}=\frac{\frac{1}{2}\cdot{}S}{2}=\frac{S}{4}=6[cm^{2}][/tex]

[tex]\boxed{7}\quad p(ABC)\| p(A_{1}B_{1}C_{1}),\quad AA_{1}\|BB_{1}\|CC_{1},\quad AA_{1}\bot{p(ABC)} \quad \triangle{ABC}\cong\triangle{A_{1}B_{1}C_{1}} \\S_{1,призма}=S_{1,(ABCA_{1}B_{1}C_{1})}=340[dm^{2}],\\ \triangle{ABC}, AB=12[dm], BC=9[dm] \\ \text{(1случай)} \quad AB\bot{BC} \Rightarrow AC=\sqrt{AB^{2}+BC^{2}}=15[dm] \\ h_{AC}=7,2[dm], S_{ABC}=\frac{AB\cdot{BC}}{2}=\frac{AC\cdot{h_{AC}}}{2} \Leftrightarrow AB\cdot{BC}=AC\cdot{h_{AC}} \rightarrow \text{ вярно равенство } \Rightarrow \exists{\triangle{АBC}} \\ S_{ABC}=54[dm^{2}]\\ S_{1,призма}=2\cdot{S_{ABC}}+P_{ABC}\cdot{h_{призма}} \Rightarrow h_{призма}=\frac{S_{1,призма}-2\cdot{S_{ABC}}}{AB+BC+AC}=\frac{340-108}{12+9+15}=\frac{232}{36}=6\frac{4}{9}[dm]\\ \phantom{q} \\ \text{(2случай)} \quad AB\not\perp{BC} \Rightarrow AC=\sqrt{AB^{2}-BC^{2}}=3\sqrt{7}[dm] \\ h_{AB}=7,2[dm], \rightarrow AC\cdot{BC}\ne AB\cdot{h_{AB}}, \quad S_{ABC}=\frac{AC\cdot{BC}}{2}=\frac{AB\cdot{h_{AB}}}{2} \rightarrow \text{ не е вярно равенство } \Rightarrow \nexists{\triangle{АBC}}[/tex]

[tex]\boxed{8}\quad B=S_{1}-S=60[cm^{2}], H=32[cm] \Rightarrow V=\frac{1}{3}\cdot{}B\cdot{}H=\frac{1}{3}\cdot{60}\cdot{32}=640[cm^{3}][/tex]

[tex]\boxed{9}\quad V=25[cm^{3}], H=3[cm], k=3,9[cm] \\ V=\frac{1}{3}\cdot{B}\cdot{H} \Rightarrow B=\frac{3\cdot{V}}{H}=25[cm^{2}] \\ B=a^{2} \Rightarrow a=5[cm] \text{ основен ръб на правилната четириъгълна пирамида} \\ S=4\cdot{a}\cdot{k}=78[cm^{2}] \\ S_{1}=S+B=78+25=103[cm^{2}][/tex]

Всичките има сгрешени отговори!