Осми разред основне школе

Површина призме - понављање градива

Решени задаци, припрема за контролни задатак.

Задаци

Текст задатака објашњених у видео лекцији:

Пр.1)   Основна ивица праве правилне призме је 6cm, а висина призме

           је 5cm. Израчунати површину те призме ако је призма:

           а) четворострана   б) тространа   в) шестострана

Пр.2)   Ако је дијагонала коцке $10\sqrt 3 cm$, израчунати површину

           те коцке.

Пр.3)   Ако је дијагонала бочне стране правилне тостране призме 8cm 

           и ако она заклапа са равни основе угао ${45^ \circ }$, израчунати

           површину те пизме.

Пр.4)   Ако је краћа дијагонала основе правилне шестостране призме

           12cm, а дијагонала бочне стране 7cm, израчунати површину

            те призме.

 

 

Пр.1)   a)

627 png

\[\begin{gathered}
a = 6cm \hfill \\
\underline {H = 5cm} \hfill \\
P = ? \hfill \\
\hfill \\
P = 2B + M \hfill \\
\hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
B = {a^2} \hfill \\
B = {6^2} \hfill \\
B = 36c{m^2} \hfill \\
\end{gathered} &{}&\begin{gathered}
M = 4aH \hfill \\
M = 4 \cdot 6 \cdot 5 \hfill \\
M = 120c{m^2} \hfill \\
\end{gathered}
\end{array} \hfill \\
\hfill \\
P = 2 \cdot 36 + 120 \hfill \\
P = 72 + 120 \hfill \\
P = 192c{m^2} \hfill \\
\end{gathered} \]

 

б)

633 png

\[\begin{gathered}
a = 6cm \hfill \\
\underline {H = 5cm} \hfill \\
P = ? \hfill \\
\hfill \\
P = 2B + M \hfill \\
\hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
B = \frac{{{a^2}\sqrt 3 }}{4} \hfill \\
B = \frac{{{6^2}\sqrt 3 }}{4} \hfill \\
B = 9\sqrt 3 c{m^2} \hfill \\
\end{gathered} &{}&\begin{gathered}
M = 3aH \hfill \\
M = 3 \cdot 6 \cdot 5 \hfill \\
M = 90c{m^2} \hfill \\
\end{gathered}
\end{array} \hfill \\
\hfill \\
P = 2 \cdot 9\sqrt 3 + 90 \hfill \\
P = 18\sqrt 3 + 90c{m^2} \hfill \\
\end{gathered} \]

 

в)

636 png

\[\begin{gathered}
a = 6cm \hfill \\
\underline {H = 5cm} \hfill \\
P = ? \hfill \\
\hfill \\
P = 2B + M \hfill \\
\hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
B = 6 \cdot \frac{{{a^2}\sqrt 3 }}{4} \hfill \\
B = 6 \cdot \frac{{{6^2}\sqrt 3 }}{4} \hfill \\
B = 54\sqrt 3 c{m^2} \hfill \\
\end{gathered} &{}&\begin{gathered}
M = 6aH \hfill \\
M = 6 \cdot 6 \cdot 5 \hfill \\
M = 180c{m^2} \hfill \\
\end{gathered}
\end{array} \hfill \\
\hfill \\
P = 2 \cdot 54\sqrt 3 + 180 \hfill \\
P = 108\sqrt 3 + 180c{m^2} \hfill \\
\end{gathered} \]

 

Пр.2)   

624 png

\[\begin{gathered}
\underline {D = 10\sqrt 3 cm} \hfill \\
P = ? \hfill \\
\hfill \\
D = a\sqrt 3 \hfill \\
10\sqrt 3 = a\sqrt 3 \hfill \\
a = 10cm \hfill \\
\hfill \\
P = 6{a^2} \hfill \\
P = 6 \cdot {10^2} \hfill \\
P = 6 \cdot 100 \hfill \\
P = 600c{m^2} \hfill \\
\end{gathered} \]

Пр.3)   

641 png

\[\begin{gathered}
d = 8cm \hfill \\
\underline {\alpha = {{45}^ \circ }} \hfill \\
P = ? \hfill \\
\end{gathered} \]

642 png

\[\begin{gathered}
a = H \hfill \\
d = a\sqrt 2 \hfill \\
8 = a\sqrt 2 \hfill \\
a = \frac{8}{{\sqrt 2 }} \cdot \frac{{\sqrt 2 }}{{\sqrt 2 }} = 4\sqrt 2 cm \hfill \\
H = 4\sqrt 2 cm \hfill \\
\hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
B = \frac{{{a^2}\sqrt 3 }}{4} \hfill \\
B = \frac{{{{\left( {4\sqrt 2 } \right)}^2}\sqrt 3 }}{4} \hfill \\
B = \frac{{16 \cdot 2\sqrt 3 }}{4} \hfill \\
B = 8\sqrt 3 c{m^2} \hfill \\
\end{gathered} &{}&\begin{gathered}
M = 3aH \hfill \\
M = 6 \cdot 4\sqrt 2 \cdot 4\sqrt 2 \hfill \\
M = 96c{m^2} \hfill \\
\end{gathered}
\end{array} \hfill \\
\hfill \\
P = 2B + M \hfill \\
P = 2 \cdot 8\sqrt 3 + 96 \hfill \\
P = 16\sqrt 3 + 96c{m^2} \hfill \\
\end{gathered} \]

Пр.4)  

643 png 

\[\begin{gathered}
{d_m} = 12cm \hfill \\
\underline {d = 7cm} \hfill \\
P = ? \hfill \\
\hfill \\
{d_m} = a\sqrt 3 \hfill \\
12 = a\sqrt 3 \hfill \\
a = \frac{{12}}{{\sqrt 3 }} \cdot \frac{{\sqrt 3 }}{{\sqrt 3 }} = 4\sqrt 3 cm \hfill \\
\hfill \\
{d^2} = {a^2} + {H^2} \hfill \\
{7^2} = {\left( {4\sqrt 3 } \right)^2} + {H^2} \hfill \\
49 = 48 + {H^2} \hfill \\
{H^2} = 1 \hfill \\
H = 1cm \hfill \\
\begin{array}{*{20}{c}}
\begin{gathered}
B = 6 \cdot \frac{{{a^2}\sqrt 3 }}{4} \hfill \\
B = 6 \cdot \frac{{{{\left( {4\sqrt 3 } \right)}^2}\sqrt 3 }}{4} \hfill \\
B = 6 \cdot \frac{{16 \cdot 3\sqrt 3 }}{4} \hfill \\
B = 72\sqrt 3 c{m^2} \hfill \\
\end{gathered} &{}&\begin{gathered}
M = 6aH \hfill \\
M = 6 \cdot 4\sqrt 3 \cdot 1 \hfill \\
M = 24\sqrt 3 c{m^2} \hfill \\
\hfill \\
\hfill \\
\hfill \\
\end{gathered}
\end{array} \hfill \\
\hfill \\
P = 2B + M \hfill \\
P = 2 \cdot 72\sqrt 3 + 24\sqrt 3 \hfill \\
P = 168\sqrt 3 c{m^2} \hfill \\
\end{gathered} \]