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\lhead{\color{blue} \Large Lecture Notes}
\chead{\color{black} \LARGE  Area}
\rhead{\large page   \ \thepage}
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\lfoot{\small   \copyright $\;$ copyright  Hidegkuti,  Powell,  2009}
\rfoot{\small Last revised: December 20, 2013}
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\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 probably easier if we know how they were derived.\bigskip

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\textbf{Definition: \ }The area of a $1$ feet by $1$ feet square is defined
to be $1\unit{ft}^{2}$ $\ $(square-feet)$.$ (Similar definitions can be
formulated with mi$^{2},$ cm$^{2},$ in$^{2},$ etc. ) \ The area of an
object, measured in $\unit{ft}^{2},$ is the number of $1\unit{ft}$ by $1%
\unit{ft}$ square needed to cover the object, cutting and pasting allowed.%
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\bigskip 

Area is not a length like perimeter. Area is always measured in $\unit{ft}%
^{2},$ $\unit{mi}^{2},$ $\unit{cm}^{2},$ $\unit{in}^{2},$ etc., and is
usually denoted by $A$.\bigskip 

\begin{center}
{\LARGE Part 1 - Rectangles\bigskip }
\end{center}

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\textbf{Theorem: }The area of a rectangle with sides $x$ and $y$ is $A=xy$. 
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\bigskip 

Proof: \ Consider rectangle with sides $3\unit{m}$ and $5\unit{m}$. \ The
area of this rectangle will be as many $\unit{m}^{2}$ as many $1\unit{m}$ by 
$1\unit{m}$ m squares are needed to cover it. Once we place this grid on the
rectangle, it is easy to see, just how many squares we need.

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We used exactly $15$ squares to cover the rectangle, and so the area is $15%
\unit{m}^{2}$.\bigskip

Mathematicians also proved that the formula is true even if the sides of the
rectangle are not integers.

It is interesting to see that we basically counted how many square meters we
have. A computation for the area that includes the units is slightly
different. Instead of counting square meters, we literally multiply meter by
meter. 
\begin{equation*}
A=ab=3\unit{m}\left( 5\unit{m}\right) =15\unit{m}^{2}
\end{equation*}

Area computation will always yield units such as $\unit{m}^{2}$ (square
meters), or $\unit{in}^{2}$ (square inches), or $\unit{mi}^{2}$ (square
miles) and so on.\pagebreak 

\textit{Example 1: }\ Find the area of a rectangle with sides $13\unit{in}$
and $7\unit{in}$.\medskip 

Solution: \ We apply the formula $A=xy$.%
\begin{equation*}
A=xy=13\unit{in}\left( 7\unit{in}\right) =91\unit{in}^{2}
\end{equation*}%
\bigskip 

\begin{center}
{\LARGE Part 2 - Triangles\bigskip }
\end{center}

The following few area formulas will demonstrate how mathematicians work: we
will use already proven results to come up with new formulas. \ We will
first consider right triangles.

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\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}{%
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\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%
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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 

Let us now consider general triangles.\medskip 

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\textbf{Theorem:} \ The area of a general triangle with sides $a,$ $b,$ $c$
and height $h$ as shown on the picture below is $A=\dfrac{ah}{2}$.\FRAME{%
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Proof: \ As before, we will use a previously obtained result. Since the
general triangle no longer has a right angle, we create it by drawing in the
altitude or height belonging to the side $a.$ Now we split our triangle into
two right triangles, and each of them is half of a rectangle.\FRAME{dtbpF}{%
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rectangle, with sides $a$ and $h$. Thus $A=\dfrac{ah}{2}$.\bigskip 

\textit{Example 3:} \ Find the area of the triangle shown on the picture
below.\FRAME{dtbpF}{1.7876in}{1.3327in}{0pt}{}{}{insert5b.bmp}{\special%
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information given. We apply the area-formula.%
\begin{equation*}
A=\dfrac{ah}{2}=\dfrac{21\unit{in}\left( 12\unit{in}\right) }{2}=\dfrac{252%
\unit{in}^{2}}{2}=126\unit{in}^{2}
\end{equation*}

\pagebreak 

\begin{center}
{\LARGE Part 3 - Parallelograms\bigskip }
\end{center}

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\textbf{Definition: }\ A parallelogram is a four sided polygon with two
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\bigskip 

It is a proven fact that the opposite sides of a parallelogram are of equal
length. \ This is not part of the definition, but it is an important
property that we need to remember. \ Also, we could prove that the diagonals
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Another important property of parallelograms is the connection between its
angles. \ In every parallelogram, the opposite angles are equal, and the two
angles along each side add up to $180^{\circ }$. \ We call two such angles
supplemental.

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property that enables us to easily compute the area of the
parallelogram.\bigskip 

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\textbf{Theorem:\ }The area of a parallelogram with sides $a,$ $b$ and
height $h$ belonging to $a$ is $A=ah$.\FRAME{dtbpF}{1.6034in}{0.966in}{0pt}{%
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\pagebreak 

Proof: We will use (surprise, surprise!) a previously proven result. If we
cut off a triangle and paste it back as show on the picture below, we obtain
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rectangle with sides $a$ and $h$.\bigskip 

\textit{Example 4}: \ Find the area of the parallelogram shown on the
picture below.\FRAME{dtbpF}{1.7772in}{0.9591in}{0pt}{}{}{insert9.bmp}{%
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"XNPEU";}}Solution: \ We apply the formula for the area of a parallelogram.%
\begin{equation*}
A=ah=5\unit{cm}\left( 3\unit{cm}\right) =15\unit{cm}^{2}
\end{equation*}

\bigskip 

\begin{center}
{\LARGE Part 4 - Trapezoids\bigskip }
\end{center}

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\textbf{Definition:} \ A \textbf{trapezoid} is a four sided polygon with one
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\bigskip 

In proving the area formula for trapezoids, we will use a property about its
angles. \ It is proven that the two angles along a side connetcting two
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\textbf{Theorem: }The area of a trapezoid, with sides and height labeled as
on the picture below, is $A=\dfrac{a+b}{2}h$.\FRAME{dtbpF}{1.6034in}{0.966in%
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\bigskip 

Proof: \ If we use two identical trapezoids, we can make a parallelogram as
shown on the picture below.\FRAME{dtbpF}{2.6411in}{0.966in}{0in}{}{}{%
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'insert12.bmp';file-properties "XNPEU";}}We already know that the area of
this parallelogram is $A=\left( a+b\right) h$. Since our trapezoid is
exactly half of the parallelogram, its area is $A=\dfrac{\left( a+b\right) h%
}{2}$.\bigskip

\textit{Example 5}: \ Find the area of the trapezoid shown on the picture
below.\FRAME{dtbpF}{2.9092in}{1.9951in}{0pt}{}{}{insert13.bmp}{\special%
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"XNPEU";}}Solution: \ We apply the formula. \ It is important to notice that
we will not need all data given. \ For the area, we only need the lengths on
the parallel sides and the height connecting them. With that data, we use
the formula $A=\dfrac{a+b}{2}h$.%
\begin{equation*}
A=\dfrac{a+b}{2}h=\dfrac{20\unit{ft}+45\unit{ft}}{2}\left( 24\unit{ft}%
\right) =\dfrac{65\unit{ft}}{2}\left( 24\unit{ft}\right) =780\unit{ft}^{2}
\end{equation*}

\pagebreak 

\begin{center}
{\LARGE Part 5 - Circles\bigskip }
\end{center}

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\textbf{Definition: \ }A circle is the set of all points in a plane that are
equidistant to a fixed point. \ That equal distance is called the radius of
the circle, that fixed point is called the center of the circle.\FRAME{dtbpF%
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\textbf{Theorem: }The area of a circle with radius $r$ is $A=\pi r^{2}$.%
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\bigskip 

At this level of mathematics, we do not have the tools necessary to prove
this. \ The proof actually uses (again) the formula for the area of a
rectangle. \ Instead of a circle, we use just half of it, and then we
multiply the result by $2$. \ The basic idea is to first find an estimation
for the area, using rectangles\FRAME{dtbpF}{1.254in}{0.8121in}{0pt}{}{}{%
insert14.bmp}{\special{language "Scientific Word";type
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'insert14.bmp';file-properties "XNPEU";}}We CAN compute the grey area since
it is composed of rectangles. \ But this is a crude underestimation of the
actual area. The trick is that the more rectangles we use, the more accurate
the approximation becomes. \ Using more and more rectangles to estimate the
area of the semi-circle, the approximation becomes better and better.\FRAME{%
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"XNPEU";}}Calculus offers tools to find a unique number these approximations
approach. This number is the area.\bigskip

\textit{Example 6:} \ Find the area of a circle of radius $7\unit{ft}$.

Solution: \ We apply the formula $A=\pi r^{2}$.%
\begin{equation*}
A=\pi r^{2}=\pi \left( 7\unit{ft}\right) ^{2}\approx 153.\,\allowbreak 94%
\unit{ft}^{2}
\end{equation*}%
\textit{Example 7:} \ Find the area of the figure shown on the picture below.%
\FRAME{dtbpF}{1.8429in}{1.5134in}{0pt}{}{}{insert16.bmp}{\special{language
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"XNPEU";}}Solution: \ The area is the sum of the areas of a rectangle and a
semi-circle. We apply the appropriate formulas and then add the results. \
The horizontal side of the rectangle is $6$ cm since we can fit exactly two
radii on it.%
\begin{eqnarray*}
A_{\text{rectangle}} &=&ab=4\unit{cm}\left( 6\unit{cm}\right) =24\unit{cm}%
^{2} \\
A_{\text{semicircle}} &=&\dfrac{\pi r^{2}}{2}=\dfrac{\pi \left( 3\unit{cm}%
\right) ^{2}}{2}\approx 14.\,\allowbreak 137\unit{cm}^{2} \\
A &=&A_{\text{rectangle}}+A_{\text{semicircle}}\approx 24\unit{cm}%
^{2}+14.\,\allowbreak 137\unit{cm}^{2}=38.\,\allowbreak 137\unit{cm}^{2}
\end{eqnarray*}

\bigskip \bigskip \bigskip 

\begin{center}
{\LARGE Practice Problems\bigskip }
\end{center}

\begin{enumerate}
\item The sides of a rectangle are given below. \ Find the area of the
rectangle. \ Include units in your answer.

\ \ \ \ \ \ \ \ \ a) \ $4\unit{in}$ and $11\unit{in}$ \ \ \ \ \ \ \ \ \ \ \
\ \ b) \ $15\unit{cm}$ and $8\unit{cm}$ \ \ \ 

\item Find the area of the right triangles shown on the picture below.\FRAME{%
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"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}{2.5676in}{1.2142in}{0pt}{}{}{insert18.bmp%
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"1";cropright "1";cropbottom "0";filename 'insert18.bmp';file-properties
"XNPEU";}}

\item Find the area of the figure shown on the picture below. \ Include
units in your answer.\FRAME{dtbpF}{1.8645in}{1.4823in}{0pt}{}{}{insert19.bmp%
}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio
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"1";cropright "1";cropbottom "0";filename 'insert19.bmp';file-properties
"XNPEU";}}

\item Find the area of a circle with radius $7\unit{in}$ long.\pagebreak
\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 a) \ $A=24\unit{m}^{2}$ \ \ \ \ \ \ \ \ b) \ $A=210\unit{cm}^{2}$

\item $\ A=159\unit{ft}^{2}$

\item $A=54\unit{ft}^{2}$

\item $A=49\pi \unit{in}^{2}\approx 153.\,\allowbreak 938\,\unit{in}^{2}$
\end{enumerate}

\vspace{0.8in}

\vspace{6in}

\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}