Oblicz objetość graniasotsłupów i stożków

[lazarz]

Witam nie umiem rozwiazać tych 2 zadań a koniec roku blisko i na poprawie beda te zadania tylko nauczyciel zmieni dane czy ktoś mógłby rozwiazać? nie piszcie postów typu sam byś zrobił itd nie wiąże przyszłości z matmą ale lepsza ocena pomoże mi sie dostac na uczelnie . Do moderatorów skan jest wstawiony bo nie widziałem jak przekazać inaczej takie zadanie

[jaffa84]

ZADANIE 1
a)

\(\displaystyle{ l=6 \sqrt{2}}\)
\(\displaystyle{ H=15}\)
\(\displaystyle{ R=5}\)

Z tw. Pitagorasa
\(\displaystyle{ L ^{2} =H ^{2}+R ^{2}}\)
\(\displaystyle{ L ^{2} =15 ^{2}+5 ^{2}=225+25=250}\)
\(\displaystyle{ L=5 \sqrt{10}}\)

Z tw. Talesa
\(\displaystyle{ \frac{l}{L} = \frac{2r}{2R}}\)
\(\displaystyle{ r= \frac{Rl}{L}}\)
\(\displaystyle{ r= \frac{5*6 \sqrt{2} }{5 \sqrt{10}} = \frac{6 \sqrt{5}}{5}}\)

Z tw. Pitagorasa
\(\displaystyle{ h ^{2} =l ^{2}-r ^{2}}\)
\(\displaystyle{ h ^{2} =(6 \sqrt{2})^{2}-( \frac{6 \sqrt{5}}{5})^{2} = \frac{360}{5}- \frac{36}{5}= \frac{324}{5}}\)
\(\displaystyle{ h= \frac{18 \sqrt{5} }{5}}\)

Objętośc:
\(\displaystyle{ V= \frac{1}{3} \Pi *r ^{2}*h}\)
\(\displaystyle{ V= \frac{1}{3} \Pi *( \frac{6 \sqrt{5}}{5})^{2}*\frac{18 \sqrt{5} }{5}}\)
\(\displaystyle{ V=\frac{6 \sqrt{5} }{5} \Pi * \frac{36}{5}= \frac{216\sqrt{5}}{25}}\)

b)

\(\displaystyle{ H=7}\), \(\displaystyle{ h=5}\), \(\displaystyle{ R=5}\)

Z tw. Talesa:
\(\displaystyle{ \frac{h}{H} = \frac{r}{R}}\)
\(\displaystyle{ r=\frac{hR}{H}= \frac{5*5}{7} = \frac{25}{7}}\)

\(\displaystyle{ V _{1}= \frac{1}{3} \Pi*r ^{2} h}\)
\(\displaystyle{ V _{2}= \frac{1}{3} \Pi*R^{2} H}\)
\(\displaystyle{ V _{3}= V _{2}-V _{1}}\)

\(\displaystyle{ V _{1}= \frac{1}{3} \Pi*(\frac{25}{7} )^{2} *5= \frac{5}{3} \Pi*\frac{625}{49}= \frac{3125}{147}\Pi}\)

\(\displaystyle{ V _{2}= \frac{1}{3} \Pi*5^{2} 7 = \frac{7*25}{3}\Pi= \frac{175}{3}\Pi}\)

\(\displaystyle{ V _{3}=\frac{175}{3}\Pi -\frac{3125}{147}\Pi =\frac{8575}{147}\Pi -\frac{3125}{147}\Pi=\frac{5450}{147}\Pi}\)


c)

\(\displaystyle{ H=15}\), \(\displaystyle{ R=5}\), \(\displaystyle{ h _{1}=10}\), \(\displaystyle{ h _{2}=3}\)

\(\displaystyle{ V _{3}=V _{1}-V _{2}}\)

\(\displaystyle{ V _{1}= \frac{1}{3} \Pi*r _{1} ^{2} h_{1}}\)
\(\displaystyle{ V _{2}= \frac{1}{3} \Pi*r _{2} ^{2} h_{2}}\)

Z tw. Talesa
\(\displaystyle{ \frac{h_{1}}{H} = \frac{r_{1}}{R}}\)
\(\displaystyle{ r_{1}=\frac{h_{1}R}{H}= \frac{10*5}{15}= \frac{10}{3}}\)

\(\displaystyle{ V _{1}= \frac{1}{3} \Pi*(\frac{10}{3}) ^{2} *10=\frac{10}{3} \Pi*\frac{100}{9}=\frac{1000}{27} \Pi}\)

Z tw. Talesa
\(\displaystyle{ \frac{h_{2}}{h_{1}} = \frac{r_{2}}{r_{1}}}\)

\(\displaystyle{ r_{2}=\frac{h_{2}r_{1}}{h_{1}}= \frac{3* \frac{10}{3} }{10}= \frac{10}{10}=1}\)

\(\displaystyle{ V _{2}= \frac{1}{3} \Pi*1 ^{2} *3=\Pi}\)

\(\displaystyle{ V _{3}=\frac{1000}{27} \Pi-\Pi=\frac{973}{27} \Pi}\)


ZADANIE 2

a)
\(\displaystyle{ A=7}\)
\(\displaystyle{ C=12}\)
\(\displaystyle{ H=10}\)
\(\displaystyle{ c=4}\)

\(\displaystyle{ V _{3}=V _{1}-V _{2}}\)

\(\displaystyle{ V _{1}= \frac{1}{2}ABH}\)
\(\displaystyle{ V _{2}= \frac{1}{2}abh}\)

Z tw. Pitagorasa:
\(\displaystyle{ B ^{2} =C^{2}-A^{2}}\)
\(\displaystyle{ B ^{2} =12^{2}-7^{2}=144-49=95}\)
\(\displaystyle{ B= \sqrt{95}}\)

\(\displaystyle{ V _{1}= \frac{1}{2}7*\sqrt{95}*10=35\sqrt{95}}\)

Z tw. Talesa:
\(\displaystyle{ \frac{b}{B} = \frac{c}{C}}\)
\(\displaystyle{ b = \frac{cB}{C}=\frac{4\sqrt{95}}{12} =\frac{\sqrt{95}}{3}}\)

Z tw. Pitagorasa:
\(\displaystyle{ a ^{2} =c^{2}-b^{2}}\)
\(\displaystyle{ a ^{2} =4^{2}-(\frac{\sqrt{95}}{3})^{2}=16- \frac{95}{9} = \frac{49}{9}}\)
\(\displaystyle{ a= \frac{7}{3}}\)

Z tw. Talesa
\(\displaystyle{ \frac{h}{H} = \frac{c}{C}}\)
\(\displaystyle{ h = \frac{cH}{C}=\frac{4*10}{12} =\frac{10}{3}}\)

\(\displaystyle{ V _{2}= \frac{1}{2}\frac{7}{3} *\frac{\sqrt{95}}{3}*\frac{10}{3}=\frac{35\sqrt{95}}{27}}\)

\(\displaystyle{ V _{3}=35\sqrt{95}-\frac{35\sqrt{95}}{27}=\frac{495\sqrt{95}}{27}-\frac{35\sqrt{95}}{27}=\frac{910\sqrt{95}}{27}}\)


b)

\(\displaystyle{ V _{3}=V _{1}-V _{2}}\)

\(\displaystyle{ V _{2}=2,5*5*10=125}\)

Graniastosłupy są podobne, więc:
\(\displaystyle{ \frac{H}{10}= \frac{5}{2,5}}\)
\(\displaystyle{ H= \frac{50}{2,5}=20}\)

\(\displaystyle{ V _{1}=5*10*20=1000}\)

\(\displaystyle{ V _{3}=1000-125=875}\)

c)
\(\displaystyle{ A=6}\)
\(\displaystyle{ B=14}\)
\(\displaystyle{ H=8}\)
\(\displaystyle{ D=10}\) - wysokośc podstawy
\(\displaystyle{ h=6}\)


\(\displaystyle{ V _{3}=V _{1}-V _{2}}\)

\(\displaystyle{ V _{1}= \frac{1}{2} (A+B)D*H}\)
\(\displaystyle{ V _{2}= \frac{1}{2} (a+b)d*h}\)

\(\displaystyle{ V _{1}= \frac{1}{2} (6+14)*10*8=20*40=800}\)

Z podobieństwa figór mamy:
\(\displaystyle{ \frac{h}{H}= \frac{a}{A} = \frac{b}{B} = \frac{d}{D}}\)

\(\displaystyle{ a= \frac{Ah}{H}= \frac{6*6}{8}= \frac{9}{2}}\)
\(\displaystyle{ b= \frac{Bh}{H}= \frac{14*6}{8}= \frac{21}{2}}\)
\(\displaystyle{ d= \frac{Dh}{H}= \frac{10*6}{8}= \frac{60}{4} =15}\)

\(\displaystyle{ V _{2}= \frac{1}{2} ( \frac{9}{2} +\frac{21}{2} )*15*6=45*15=675}\)

\(\displaystyle{ V _{3}=800-675=125}\)

POWODZENIA NA SPRAWDZIANIE