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Area of a Trapezium Theory
![\begin{tabular}{p{8cm}c} \begin{tabular}{p{8cm}} Area of trapezium.\newline \(A_1=\dfrac{1}{2} b h,\; A_2=a h,\; A_3=\dfrac{1}{2} ch\)\newline \(\begin{aligned} \text { Total area } & =\frac{1}{2} b h+a h+\frac{1}{2} c h \\ & =\frac{1}{2} h(b+2 a+c) \\ & =\frac{1}{2} h(a+b+c+a) \end{aligned}\)\newline \(\begin{aligned} &\text { Let } d =a+b+c \\ &\therefore \text { Area } =\frac{1}{2} h(d+a) . \end{aligned}\) \end{tabular} & \begin{tabular}{c} \begin{tikzpicture}[scale=1.2] \coordinate[label=below left:] (O) at (0,0); \coordinate[label=above left:] (A) at (70:2cm); \begin{scope}[shift=(A)] \coordinate[label=above right:] (B) at (right:2cm); \end{scope} \begin{scope}[shift=(O)] \coordinate[label=below right:] (C) at (right:4cm); \end{scope} \coordinate[label=below:] (D) at ($(O)!(A)!(C)$); \coordinate[label=below:] (E) at ($(O)!(B)!(C)$); \draw[line width=1pt] (O)--(A) node[midway,sloped]{}--(B)node[sloped,midway]{>}--(C)node[midway,sloped]{}--(O)node[sloped,midway]{>}; \draw[dashed,line width=1pt] (A)--(D) node[right,midway] {\(h\)}; \draw[dashed,line width=1pt] (B)--(E); \node (F) at (45:0.6cm) {\(A_1\)}; \begin{scope}[shift=(F)] \node at (right:1.3cm) {\(A_2\)}; \end{scope} \begin{scope}[shift=(F)] \node at (right:2.8cm) {\(A_3\)}; \path (O)--(D) node[below,midway] {\(b\)}--(E) node[below,midway] {\(a\)}--(C) node[below,midway] {\(c\)}; \path (A)--(B) node[above,midway] {\(a\)}; \end{scope} \end{tikzpicture} \end{tabular} \end{tabular} \begin{multicols}{2} \textbf{Example 1}\\ Find the area of a trapezium with a height of \(6 \text{~cm}\) and the lengths of the parallel sides \(4 \text{~cm}\) and \(8 \text{~cm}\).\\ \textbf{Example 1 solution}\\ \(\begin{aligned} \text { Area } & =\frac{1}{2} h(d+a) \\ & =\frac{1}{2} \times 6(8+4) \\ & =3 \times 12 \\ & =36 \text{~cm}^2 \end{aligned}\) \columnbreak \textbf{Example 2}\\ The area of a trapezium is \(60 \text{~cm}^2\). The lengths of the parallel sides are \(10\text{~cm}\) and \(5\text{~cm}\), find the height.\\ \textbf{Example 2 solution}\\ \(\begin{aligned} \text { Area } & =\frac{1}{2} h(d+a) \\ 60 & =\frac{1}{2} h(10+5) \\ 2 \times 60 & =h \times 15 \\ 15 h & =2 \times 60 \\ h & =\frac{2 \times 60}{15} \\ h & =2 \times 4 \\ & =8 \mathrm{~cm}\end{aligned}\) \end{multicols}](/media/v53j2mo1/7956.png)
NSW Y8 Maths Area and Perimeter Area of a Trapezium
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