NSW Y8 Maths Volume Volume of a Prism

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Volume of a Prism Theory

\begin{minipage}[c]{6cm} The volume of a prism is given by  $$V= \text{ Base area } \times \text{ height}.$$ \end{minipage}\\  \begin{multicols}{2} \textbf{Example 1}\\ Find the volume of the triangular prism.\\ \begin{center} \begin{tikzpicture}[scale=0.4,line width=1pt] \coordinate[label=above:] (O) at (0,0); \coordinate[label=right:] (B) at (90:4); \coordinate[label=left:] (A) at (right:5); \begin{scope}[shift=(B)] \coordinate[label=left:] (C) at (45:6); \end{scope} \begin{scope}[shift=(A)] \coordinate[label=left:] (D) at (45:6); \end{scope} \draw (B)--(O) node[left,midway]{4 cm}--(A)node[below,midway]{5 cm}--cycle; \draw (B)--(C)--(D)--(A) node[right,midway] {6 cm}; \pic [draw=black,line width=1pt,angle radius=0.4cm,angle eccentricity=1.6,""] {right angle=A--O--B}; \end{tikzpicture} \end{center} \textbf{Example 1 solution}\\ \(\begin{aligned} \text { Area of base } & =\frac{1}{2} \times 4 \times 5 \\ & =10 \text{~cm}^2  \end{aligned}\)\\ \(\begin{aligned} \therefore \text { Volume } & =10 \times 6 \\ & =60 \text{~cm}^3 \end{aligned}\)  \columnbreak \textbf{Example 2}\\ Find the volume of the prism that has a trapezium base.\\ \begin{center} \begin{tikzpicture}[scale=0.55,line width=1pt] \coordinate[label=above:] (O) at (0,0); \coordinate[label=right:] (B) at (90:4); \coordinate[label=left:] (A) at (right:5); \begin{scope}[shift=(B)] \coordinate[label=left:] (C) at (right:5); \end{scope} \begin{scope}[shift=(B)] \coordinate[label=left:] (D) at (60:3); \end{scope} \begin{scope}[shift=(D)] \coordinate[label=left:] (F) at (right:2); \end{scope} \begin{scope}[shift=(C)] \coordinate[label=left:] (E) at (120:2); \end{scope} \coordinate[label=left:] (G) at (60:3); \begin{scope}[shift=(G)] \coordinate[label=left:] (H) at (right:2); \end{scope} \coordinate[label=left:] (I) at ($(G)!0.5!(H)$); \coordinate[label=left:] (J) at ($(O)!0.5!(A)$); \draw (O)--(B) node[left,midway]{4 cm}--(C)--(A)--node[below,midway]{5 cm} cycle; \draw (B)--(D)--(F)node[above,midway]{2 cm}--(C); \draw[dashed] (O)--(G) edge (D)--(H) edge (F)--(A); \draw[latex-latex] (I)--(J)node[rectangle, fill=gray!10!white,midway]{3 cm}; \end{tikzpicture} \end{center}  \textbf{Example 2 solution}\\ \(\begin{aligned} \text { Area of base } & =\frac{1}{2} \times 3(2+5) \\ & =\frac{1}{2} \times 3 \times 7 \\ & =\frac{21}{2} \text{~cm}^2  \end{aligned}\)\\ \(\begin{aligned} \text { Volume } & =\frac{21}{2} \times 4 \\ & =42 \text{~cm}^2 \end{aligned}\) \end{multicols}

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