\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.4in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \large Lecture Notes} \chead{\color{black} \Large Area of Right Triangles} \rhead{ page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ Hidegkuti, Powell, 2009} \rfoot{\small Last revised: September 14, 2016} \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{29pt} \setlength{\parindent}{0in} \begin{document} The \textbf{area} of a geometric object is a measurement of its surface.\bigskip While we could think about perimeter as a fencing problem, area can be thought of as follows. Suppose a geometric object is a room. How much rug do we need to buy to cover the entire room? \ Understanding and remembering the area formulas are easier if we know how they were derived.\bigskip \bigskip Recall that the area of a rectangle with sides $x$ and $y$ is $A=xy$. \bigskip %TCIMACRO{% %\TeXButton{box start - 7.5in}{\begin{boxedminipage}{7.5in} %\setlength{\fboxrule}{15pt} %\setlength{\fboxsep}{15pt}}}% %BeginExpansion \begin{boxedminipage}{7.5in} \setlength{\fboxrule}{15pt} \setlength{\fboxsep}{15pt}% %EndExpansion %TCIMACRO{\TeXButton{white}{\color{white}}}% %BeginExpansion \color{white}% %EndExpansion .% %TCIMACRO{\TeXButton{black}{\color{black}}}% %BeginExpansion \color{black}% %EndExpansion \textbf{Theorem: }The area of a right triangle with sides $a$, $b$, and $c$ (where $c$ is the longest side) is $A=\dfrac{ab}{2}$. \ \FRAME{dtbpF}{% 2.0384in}{1.0741in}{0pt}{}{}{insert3.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.0384in;height 1.0741in;depth 0pt;original-width 2.0003in;original-height 1.0395in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert3.bmp';file-properties "XNPEU";}} %TCIMACRO{\TeXButton{end of box}{\end{boxedminipage}}}% %BeginExpansion \end{boxedminipage}% %EndExpansion \bigskip Proof: \ It is very easy to see that every right triangle is basically half of a rectangle. We can make a rectangle if we use two identical right triangles as shown on the picture below.\FRAME{dtbpF}{2.4206in}{1.2816in}{0pt% }{}{}{insert4.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.4206in;height 1.2816in;depth 0pt;original-width 2.38in;original-height 1.2462in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert4.bmp';file-properties "XNPEU";}}Since the rectangle's sides are $a$ and $b$, its area is $ab.$ The area of our triangle must be half of it. Thus $A=\dfrac{ab}{2}$. Notice that we never used the length of the longest side, $c$.\bigskip \textit{Example 2:} \ Find the area of the right triangle with sides $5\unit{% m}$, $12\unit{m}$, and $13\unit{m}$ long.\medskip Solution: \ It is important to know that the largest side, $13\unit{m}$ long, is not needed for this computation. With labeling $a=5\unit{m}$ and $% b=12\unit{m}$, the area is% \begin{equation*} A=\dfrac{ab}{2}=\dfrac{5\unit{m}\left( 12\unit{m}\right) }{2}=\dfrac{60\unit{% m}^{2}}{2}=30\unit{m}^{2} \end{equation*} \pagebreak \begin{center} {\LARGE Practice Problems\bigskip } \end{center} \begin{enumerate} \item Find the area of the right triangles shown on the picture below.\FRAME{% dtbpF}{4.6458in}{1.5567in}{0pt}{}{}{insert17.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 4.6458in;height 1.5567in;depth 0pt;original-width 4.5939in;original-height 1.5203in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'insert17.bmp';file-properties "XNPEU";}} \item Find the area of the figure shown on the picture below. \ Include units in your answer.\FRAME{dtbpF}{3.1107in}{1.4148in}{0pt}{}{}{pic1.bmp}{% \special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 3.1107in;height 1.4148in;depth 0pt;original-width 3.0666in;original-height 1.3802in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic1.bmp';file-properties "XNPEU";}} \item Find the value of $x$ based on the picture if we know that the area of the object on the picture is $64\unit{ft}^{2}$.\FRAME{dtbpF}{2.7423in}{% 1.4148in}{0pt}{}{}{pic2.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.7423in;height 1.4148in;depth 0pt;original-width 2.7in;original-height 1.3802in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename 'pic2.bmp';file-properties "XNPEU";}} \bigskip \bigskip \bigskip \end{enumerate} \begin{center} {\Large Practice Problems - Answers}\bigskip \end{center} \begin{enumerate} \item a) \ \ $A=44\unit{in}^{2}$ \ \ \ \ \ \ \ b) \ $A=60\unit{in}^{2}$ \item $199\unit{ft}^{2}$ \item $x=4$ \end{enumerate} \bigskip \href{http://www.teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html% }{For more documents like this, visit our page at\ http://www.teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}