Resources for Area of a Trapezium
-
Questions
10
With Worked SolutionClick Here -
Video Tutorials
2
Click Here
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
Videos
Videos relating to Area of a Trapezium.
-
Area of a Trapezium - Video - Area of a Trapezium 1
-
Area of a Trapezium - Video - Area of a Trapezium 2
Plans & Pricing
With all subscriptions, you will receive the below benefits and unlock all answers and fully worked solutions.
-
Teachers TutorsFeaturesFreeProAll ContentAll courses, all topicsQuestionsAnswersWorked SolutionsSystemYour own personal portalQuizbuilderClass ResultsStudent ResultsExam RevisionRevision by TopicPractise ExamsAnswersWorked SolutionsAwesome StudentsFeaturesFreeProContentAny course, any topicQuestionsAnswersWorked SolutionsSystemYour own personal portalBasic ResultsAnalyticsStudy RecommendationsExam RevisionRevision by TopicPractise ExamsAnswersWorked Solutions