\tableofcontents

\section{How to Use This Guide}
The \textit{Self-Paced Review} consists of review modules with exercises; problems and solutions; self-tests and solutions; and self-evaluations for the four topic areas Algebra, Geometry and Analytic Geometry, Trigonometry, and Exponentials \& Logarithms. In addition, previous \textit{Diagnostic Exams} with solutions are included. Each topic area is independent of the others. 

The \textit{Review Modules} are designed to introduce the core material for each topic area. A numbering system facilitates easy tracking of subject material. For example, in Algebra, the subtopic Linear Equations is numbered with \ref{alg:ex:two:three}. Problems and self-evaluations are categorized using this numbering system.

When using the \textit{Self-Paced Review}, it is important to differentiate between concept learning and problem solving. The review modules are oriented toward refreshing concept understanding while the problems and self-tests are designed to develop problems solving ability. When reviewing the modules, exercises are liberally sprinkled throughout the modules: solve these exercises when working through the module. The problems should be attempted without looking at the solutions. If a problem cannot be solved after at least two honest efforts, then consult the solutions. Trying many times and then succeeding results in a better understanding than trying several times and reading the solution. 

The tests should be taken only when both understanding of the material and problem solving ability have been achieved. The self-evaluation is a useful tool to evaluate the mastery of the material. Finally, the previous Diagnostic Exams should provide the finishing touch.

The review modules were written by Professor A\. P\. French (Physics Department) and Adeliada Moranescu (MIT Class of 1994). The problems and solutions were written by Professor Arthur Mattuck (Mathematics Department). This document was originally produced by the Undergraduate Academic Affairs Office, August, 1992, and transcribed to \LaTeX\ and edited for OCW by Tea Dorminy (MIT Class of 2013) in August, 2010.

\section{Geometry Review Module}
This is a short module. It is not intended to be a full review of the theorems of Euclidean geometry. Its purpose is just to remind you of some of the chief results that will be of direct use to you in your work in science and other areas of mathematics. In particular, thelane geometry in this module leads quickly into trigonomoetry, which is the subject of a sepparate module.

\subsection{Triangles}

You will be familiar with the fact that, in Eucilidean geometry, the angles of a triangle add up to $180^\circ$. We shall take this property of triangles as a given. Itis worth remembering, though, that this is not a general result. It holds only for \textit{[plane} triangles. The angles of a triangle drawn on the surface of a sphere add up to \textit{more} than $180^\circ$. The reason whi that isn't important to us for most purposes is that the triangles we deal with usually have very small linear dimensions relative to the radius of the Earth, so, for example, most usrveying on the earth's surface can use Euclidean geometry with no significant error. (And then there are also other ``non-Euclidean geometries'' that obey different rules, whicfh we'll certainly not deal with.) But here, this $180^\circ$ property is a starting point:

In a triangle $ABC$, \[\angle{A}+\angle{B}+\angle{C}=180^\circ\]

\begin{wrapfigure}{r}{1.5in}
\begin{asy}
size(1.5inch);
draw((0,0)--(4,4)--(6,0)--(0,0));
draw((6,0)--(8,0));
label("$B$",(0,0),SW);
label("$C$",(6,0),S);
label("$A$",(4,4),N);
label("$C''$",(6,0),NE);
label("$D$",(8,0));
\end{asy}
\end{wrapfigure}

Another familiar property is that, if one side of a triangle is extended, as in the diagram here, the exterior angle $C''$ is equal to the sum of the interior angles $A$ and $B$. (A reminder of the proof: Since $BCD$ is a straight line, $\angle{C}+\angle{C''}=180^\circ$, so $\angle{C''}=180^\circ-\angle{C}$. Then use the result above.)

\subsubsection{Angles and Similar Triangles}
TODO diagram

Some basic results of Euclidean geometry are very useful in analyzing geometrical situations in general and triangles in particular. When one straight line cuts across two parallel straight lines:
\begin{itemize}
\item $\angle{A}=\angle{B}$ (because lines are parallel)
\item $\angle{B}=\angle{C}$ (opposite angles are equal)
\item $\angle{A}=\angle{C}$ (alternate interior angles are equal)
\end{itemize}

These find many applications in recognizing and relating \textit{similar} tirangles. It is easy to recognize similar triangles when they are placed side by side in the same orientation, like those on the right. It may not be so easy if the triangles are in quite different orientations, like those on the right.

TODO diagram

A common type of situation is the following mechanics problem:

A block sits on an inclined plane. It is acted on by the vertical force of gravity. In the solution fo the problem, you want to resolve this gravitational force (the weight $W$) into components parallel and perpindicular to the inclined plane. The triangle $LMN$ representing this analysis of the downard force $W$ into two mutually perpendicular parts, $F_1$ and $F_2$, is similar to the tirangle $ABC$ composed of the inclined plane and its horizontal and vertical dimensions. One can therefore put:
\[F_1:F_2:W=BC:AC:AB\]
(This is more than an exercise in pure geometry, because we are comparing a geometrical triangle to a triangle of forces. BUt it is important and useful to know that this can be done.)

TODO diagram

\begin{exercise}
Some artists used to use a \textit{camera obscura} (literally just a ``dark room'') with a small hole in one wall to form an image of a distance scene on a screen inside the room. (Then all they had to do was trace over the image!) Imagine a situation when the screen was 2 meters from the hole, forming an image of a building 10 meters high and 50 meters away. Draw a sketch showing rays of light passing to the screen, through the point-hole, from the top and bottom of the building, and calculate the height of the image.
\end{exercise}

\begin{exercise}
Find three similar triangles in the figure opposite, and prove that $s^2=rt$.

TODO figure
\end{exercise}

\begin{exercise}
In the example of an inclined plane idscussed above, find the forces $F_1$ and $F_2$ in terms of $W$ if $\angle{A}=30^\circ$.
\end{exercise}

Note: The answers to the exercises are all collected together at the end of this module. We have tried to eliminate errors, but if you find anything that you think needs to be corrected, please write to us.

\subsubsection{Orthocenters and Centroids}
In any triangle:

The three perpindicular lines from the angles to the opposite sides (the altitudes) intersect at a common point. This is the \textit{orthocenter}. 

TODO figure

The tree lines from the angles to the midpoints of the opposite sides (the medians) intersect at a single point. This is the \textit{centroid}.

TODO figure

The centroid is of particular significance physically. If a triangle is made  from a sheet of material of uniform thickness, then the centroid is its balance point (its center of gravity), $G$. This can be understood by drawing lines parallel to each of the sides in turn. SUppose the figure is supsened from the corner $A$. Then the center of each strip lies on the line $AM$, and the triangle will be in balance, when hung from $A$, with $AM$ vertical. Similarly for the other two medians. So, if suspended from the centroid, the triangle has no tendency to take up any particular orientation.

\subsection{Some Other Plane Figures}

\subsubsection{Quadrilaterals}
TODO figure

For any quadrilaterla, the sum of the angles is $360^\circ$:
\[A+B+C+D=360^\circ\]

(Think of it as two triangles, $ABC$ and $BCD$.)

There are some special quadrilaterals:

TODO figure

Trapezoids have $AB$ parallel to $CD$. Its width at half altitude is equal to the average of $AB$ and $CD$.

TODO figure

Parallelograms have $AB$ parellel to $CD$, and $AD$ parellel to $BC$. Other properties:
\begin{tabular}{cc}
$\angle{A}=\angle{C}$&$\angle{B}=\angle{D}$\\
$AB=DC$&$AD=BC$
\end{tabular}

TODO figure

Rhombus: a parallelogram with all sides equal:
\[AB=BC=CD=AD\]
Its diagonals are perpindicular.

\subsubsection{Other Polygons}
The literal meaning of `polygon' is ``many-cornered'', but we usually think in terms of the number of sides. Since the number of angles equals the number of sides, I$n$, you can't go wrong. It's useful to know the names of a few polygons beyond the quadrilateral:

\begin{tabular}{cc}
$n=5$&Pentagon\\
$n=6$&Hexagon\\
$n=7$&Heptagon\\
$n=8$&Octagon\\
$n=10$&Decagon\\
$n=12$&Dodecagon
\end{tabular}

(The ones omitted are seldom met.)

There is a general formula for the sum of all the interior angles of a polygon with $n$ sides or corners: $(n-2)\times 180^\circ$. (Note that you can obtain this by dividing the polygon up into triangles in any fashion, and noting that the sum of the interior angles of the polygon is the sum of all the interior angles of all the triangles that the polygon is divided into.)

\subsection{The Almighty Circle}
Some properties worth knowing about, even if you don't memorize them:

TODO diagram

Given any arc of the circle, the angle it subtends at the center is twice the angle it subtends at the circumference. (Special case: A semicircle subtends $180^\circ$ at the center and a right angle at the circumference.)

TODO diagram

Intersecting chords: The products of their separate parts are equal:\[ab=cd\] This is also known as ``Power of a Point.'' (Special case: A diameter and a chord perpindicular to it, with the figure labeled as shown, $y(2R-y)=x^2$. 

TODO diagram

\textit{If $y\ll R$, then $y\approx x^2/2R$.} This means that \textit{a small part of a circle has almost the exact same chape as a parabola.} That is important in many mathematical and physical approximations. Admittedly, this is analytic geometry, not plain geometry!)

The triangles formed by any two intersecting chords, not necessarily passing through the center of the circle, are similar. TODO diagram

\subsection{Lengths, Areas, \& Volumes}

There are many useful formulas for the perimeters, areas, and volumes of geometrical figures. Some of them you should definitely know. But before you look at the tabulation of important ones, consider the following statements concerning the \textit{dimensions} of such quantities:

\begin{itemize}
\item Any \textit{perimeter} must have the dimensions of \textit{length}. That is, it must be expressed in terms of units of length to the first power only.
\item Any \textit{area} must have dimensions of \textit{(length)}${}^2$.
\item Any \textit{volume} must have dimensions of \textit{(length)}${}^3$.
\end{itemize}

For example, if you are trying to remember the formula for the \textit{area} of a \textit{circle}, consider that it couldn't possibly be $2\pi r$. Whatever else there may be, it must have $r$ to the 2nd power. Guessing that it's $2\pi r^2$ (as opposed to the correct formula $\pi r^2$) is an error you shouldn't make --- but it's less serious than using a formula that could only represent a length.

Likewise, the circumference of a circle could not possibly be given by $\pi r^2$ or $2\pi r^2$.

TODO formula table

\begin{exercise}
Prove that the area of a flat circular ring or washer, with inner radius $r_1$ and outer radius $r_2$, is equal to $2\pi r_{avg}\Delta r$, where $r_{ave}$ is the average radius $\frac{r_1+r_2}2$ and $\Delta r$ is the width $r_2-r_2$.
\end{exercise}

\begin{exercise}
Given that a sphere has surface area $64\pi$, what is its volume?
\end{exercise}

\begin{exercise}
A cube of edge-length $L$ has a sphere just fitting inside it (i.e. the diameter of the sphere is equal to $L$). Calculate the ratios of the surface areas and of the volumes of these two figures.
\end{exercise}

\begin{exercise}
A plane, parallel to the base of a circular cone of height $h$ and of radius $R$ at the base, cuts across it at a distance of $h/3$ from the top. Calculate the ratios of the volumes of the small cone and the original cone.
\end{exercise}

\begin{exercise}
A cube of side $a$ is packed full with small spheres of diameter $a/n$ ($n$ being a positive integer), as shown in the cross-section diagram at right. Consider the total volume of the spheres. Does it increase or decrease as $n$ gets bigger?

Given two equal-size and equal-price boxes of mothballs, should you buy the one with the larger mothballs or the smaller mothballs?

What if mothballs' effectiveness is determined by their surface area, not their volume?

TODO diagram
\end{exercise}

\section{Answers to Exercises}

Exercise \ref{geo:ex:two:one:one}: TODO diagram

Diagram not to scale. 

Find similar traingles (see diagram above):
	\[\triangle ABC\sim \triangle CDF \rightarrow \frac{AB}{BC}=\frac{DF}{CD}\]

or, using the distacnes given, $\frac{5}{50}=\frac{DF}{2}\rightarrow DF=\frac 15$. Then the height FE of the image is $FE=2\cdot DF=\frac 25 m=0.4m$.

Note: If the artsits used only one hole in the wall, they would trace a \textit{inverted} image on the screen. (The rays passing through the hole form the top and the bottom of the building end up on the screen at points $F$ and $E$, respectively.)

TODO digram

In order to get the image right side up we would need two holes and two screens (see diagram at left).

Exercise \ref{geo:ex:two:one:two}: TODO diagram

\begin{align}
\triangle ABC \sim \triangle BDA &\rightarrow \frac{AB}{AC}=\frac{BD}{AD}\\
\triangle ABC \sim \triangle ADC &\rightarrow \frac{AB}{AC}=\frac{AD}{DC}
\end{align}

Using both (1) and (2), we have \[\frac{BD}{AD}=\frac{AD}{DC}\rightarrow \frac rs=\frac st\rightarrow s^2=rt\] TODO diagram

Solution without using similar triangles, but using the Pythagorean theorem instead:

\begin{align}
AB^2=BD^2+AD^2&\rightarrow AB^2=r^2+s^2\\
AC^2=AD^2+DC^2&\rightarrow AC^2=s^2+t^2
\end{align}

But $BC^2=AB^2+AC^2$, and $BC=r+t$; by substitution into above, we obtain:
\begin{align*}
BC^2&=r^2+s62+s^2+t^2\\
(r+t)^2&=r^2+2s^2+t^2\\
r^2+2rt+t^2=r^2+2s^2+t^2
\end{align*}

or $2rt=s^2$, which gives us the answer $rt=s^2$.

Exercise \ref{geo:ex:two:one:three}: TODO diagram

$\triangle ABC\sim \triangle LMN \rightarrow \frac{AB}{LM}=\frac{AC}{MN}=\frac{BC}{LN}$, or $\frac{AB}{W}=\frac{AC}{F_2}=\frac{BC}{F_1}$, thus $\frac{BC}{AB}=\frac{F_1}{W}$ and $\frac{AC}{AB}=\frac{F_2}{W}$.

We have $\angle A=30^\circ$. Using the known relationships in a 30-60-90 triangle, mentioned in the Trigonometry Module, $\frac{BC}{AB}=\frac 12$, $\frac{AC}{AB}=\frac{\sqrt{3}}2$

Using the equalities above, we obtain:

\[\frac{BC}{AB}=\frac 12=\frac{F_1}{W}\rightarrow F_1=\frac{W}{2}\]
\[\frac{AC}{AB}=\frac{\sqrt{3}}2=\frac{F_2}{W}\rightarrow F_2=\frac{W\sqrt{3}}2\]

Exercise \ref{geo:ex:two:four:one}:

TODO diagrams.

$A_{outer}=\pi r_2^2$, and $A_{inner}=\pi r_1^2$.

The washer is what is left of the outer circle, when the inner one is taken away.

Its area will be:

$A_w=\pi r_2^2-\pi r_1^2$

Using a few algebraic manipulations, 

$A_w=\pi(r_2^2-r_1^2)=\pi(r_2+r_1)(r_2-r_1)=2\pi\frac{r_2+r_1}{2}\left(r_2-r_1\right)=2\pi r_{avg}\Delta r$

Exercise \ref{geo:ex:two:four:two}:

TODO diagram

$A_{sphere}=4\pi R^2=64\pi$

$R^2=16, R=4$

$V_{sphere}=\frac{4\pi R^3}{3}=\frac{4\pi \cdot 64}{3}=\frac{256\pi}{3}$

Exercise \ref{geo:ex:two:four:three}:

TODO diagrams.

The radius of the sphere is $R=\frac L2$.

For the first part, the area of the cube is $6\cdot L^2$, and the surface area of the sphere is $4\pi R^2=4\pi\left(\frac L2\right)^2=\pi L^2$. Thus the ratio is
\[\frac{A_{cube}}{A_{sphere}}=\frac{6L^2}{\pi L^2}=\frac 6\pi\approx 1.91\]

For the second part, $V_{cube}=L^3$. $V_{sphere}=\frac{4\pi R^3}{3}=\frac{4\pi}{3}\left(\frac L2\right)^3=\frac{\pi L^3}{6}$. Thus the ratio is
\[\frac{V_{cube}}{V_{sphere}}=\frac{L^2}{\frac{\pi L^3}{6}}=\frac 6\pi \approx 1.91\]

Exercise \ref{geo:ex:two:four:four}:

TODO diagram

\[\triangle ABC\sim \triangle ADE\rightarrow \frac{AB}{AD}=\frac{BC}{DE}\] or $\frac{h/3}{h}=\frac{BC}{R}\rightarrow BC=R/3$.

$V_{sm}=\frac{1}{3}\pi\cdot(BC)^2\cdot AB=\frac 13\pi\left(\frac R3\right)^2\cdot\left(\frac h3\right)=\frac{\pi R^2 h}{81}$

$V_{lg}=\frac 13 \pi(DE)^2\cdot AD=\frac 13 \pi R^2h$.

\[\frac{V_{sm}}{V_{lg}}=\frac{\pi R^2 h}{81}\cdot\frac{1}{\frac{\pi R^2 h}{3}}=\frac{1}{27}\]

N.\ B.: no units, since it is a \textit{ratio} of two quantities with the same dimension, (length)${}^3$. 

Another method, dimensional analysis: since volume has dimension (length)${}^3$, the formula for $V$ must be $ch^3$ for some constant $c$ related to the angle at the vertex. Thus, in the ratio, the constant cancels out, leaving $\left(\frac{h/3}{h}\right)^3=\frac{1}{27}$.

Exercise \ref{geo:ex:two:four:five}: TODO diagram

Since the diameter of the small spheres is $a/n$, the radius will be \[R=\frac{a}{2n}\] Considering the threefold symmetry of the sphere and cube, it should be clear that we are dealing with $n^3$ small spheres packed inside the cube. Take firs just one big mothball (sphere) inside the cube: $n=1,r^3=1$. Then $R_1=\frac{a}{2}$ TODO diagram

The next number of spheres we can pack is 8, with $n=2$, we get $R_2=\frac{a}{4}$, and so on: you can continue with $n=3$, packing $n^3=27$ spheres of radius $R_3=\frac{a}{6}$, etc.

The volume of one sphere is $V=\frac{4\pi R^3}{3}=\frac{4\pi}{3}\left(\frac{a}{2n}\right)^3=\frac{\pi a^3}{6n^3}$

But since we have $n^3$ spheres, the total volume of the spheres is $V_tot=n^3\cdot V=n^3\cdot\frac{\pi a^3}{6n^3}=\frac{\pi a^3}{6}$

So the final result is independent of $n$: no matter which you buy, the total mass of the mothballs -- assuming constant density -- is the same. This result is exactly the same as in Exercise \ref{geo:ex:two:four:three}: the ratio of the volume of the inscribed sphere to the volume of the cube is $\pi/6$ no matter the sidelength $L$.

If mothballs' efficacy relates to their surface area, however, the surface area of one mothball is $4\pi R^2=4\pi\left(\frac{a}{n}\right)^2=\frac{4\pi a^2}{n^2}$; so $n^3$ mothballs have surface area of $4\pi a^2 n$. This means the surface area goes up with $n$ going up, so optimally the mothballs would be as powdered as possible. (But with sufficiently small mothballs, air no longer flows over all the surface area, but essentially only the top of the pile, negating the advantage. For our approximation of efficiency, though, finer is better.)

\section{Review Problems in Geometry}

\subsection{Triangles, similarity of figures}

Two triangles are similar when the three angles of one are the same as the three angles of the other. (In practice, you only have to show this for two angles, since the third angles will then automatically be equal.)

If two triangles are similar, \textit{corresponding sides are proportional, and the altitudes on those sides are proportional}. This is the essential fact that is most often used in scientific problems.

TODO diagrams of similarity

In working the problems below, in your diagram mark the equal angles, for similar triangles, mark two of the three pairs of equal angles, as is done above. Parallel lines are indicated by arrows. Figures are not drawn to scale!

\begin{problem}
In each of the figures, tell how many degrees angle $x$ has.

TODO 3 figures.
\end{problem}

\begin{problem}
A line segment of length 3 joins two sides of a triangle, is parallel to the third side, and has distance 1 from that side and 2 from the opposite vertex. How long is the third side of the triangle?
\end{problem}

\begin{problem}
The light from a building goes through a tiny hole 100 meters away. THe image of the building is upside down on a vertical screen 2 meters away from the hole. If the image is 1.2 meters high, how tall is the building?
\end{problem}

\begin{problem}
In each of the diagrams, find the length of the line segment $x$ marked.

TODO 2 figures.
\end{problem}

\begin{problem}
In each picture, find the length of the line segment $x$ marked.

TODO 3 figures.
\end{problem}

\begin{problem}
In the accompanying figures, the smaller triangle has area $A$. What is the area of the larger triangle?

TODO 2 figures

\end{problem}

\begin{problem}
A streetlamp 20 feet high is 15 feet horizontally distant from a woman 5 feet high. How long is the shadow she casts?
\end{problem}

\begin{problem}
How long is the diagonal of
\begin{enumerate}
\item a cube of side 1
\item a rectangular box of sides 1, 2, and 3?
\end{enumerate}
\end{problem}

\subsection{Circle Problems}

\begin{problem}
Some arcs are given. Determine the angles in the left triangle.

TODO 2 figures.
\end{problem}

\begin{problem}
The smaller triangle has area $A$; what is the area of the larger triangle?

TODO figure
\end{problem}

\begin{problem}
Find the area of the kite-shaped region pictured.

TODO diagram.
\end{problem}

\begin{problem}
Find the lengths $x$ and $y$ in the accompanying diagram.

TODO diagram
\end{problem}

\begin{problem}
The circle has radius 1, the line segment $PA$ has length 2, and is tangent to the circle. What is the length of $PB$?

TODO diagram
\end{problem}

\begin{problem}
The two circles are concentric, with radii respectively 1 and 3. The chord is tangent to the inner circle. How long is it?

TODO diagram
\end{problem}

\begin{problem}
The circle has radius 2; $PA$ and $PB$ are tangents, and $PO$ has length 4. What is the area of quadrilateral $OAPB$?

TODO diagram
\end{problem}

\begin{problem}
The two line segments marked are equal, and $PA,PB$ are tangents. What are the degrees of the two circular arcs $\arck{ADB}$ and $\arck{ACB}$ having $A$ and $B$ as their endpoints?

TODO diagram
\end{problem}

\begin{problem}
Find the length of $x$ in the accompanying picture. The arrows indicate parallel line segments; $AB=1$ and $CD$ is tangent; the circle has radius 2.

TODO diagram
\end{problem}

\subsection{Areas and Volumes}

\begin{problem}
A circle has radius $R$; what is the area and perimeter of a $72^\circ$ circular sector?
\end{problem}

\begin{problem}
What is the area of a triangle having 

\begin{enumerate}
\item all sides of length 3;
\item two sides of length 4, with a $30^\circ$ angle between them?
\end{enumerate}
\end{problem}

\begin{problem}
Find the areas of the two parallelograms shown:

TODO 2 diagrams
\end{problem}

\begin{problem}
A trapezoid has parallel sides of lengths 5 and 9, and its other two sides both have length 3. What is its area?
\end{problem}

\begin{problem}
A square and a circle have the same perimeter. WHat is the ratio of their respective areas?
\end{problem}

\begin{problem}
The ring-shaped region between two concentric circles has inner radius $a$ and outer radius $b$. Give an expression for its area $A$, and show that if we draw the median circle on the ring, halfway between the inner and outer boundaries, then \[A=(\text{length of median circle})(\text{thickness of ring})\]

TODO target diag.
\end{problem}

\begin{problem}
The radius of a riht circular cylinder is $1/3$ of its height; the volume of the cylinder is $24\pi$. What is the radius and height?
\end{problem}

\begin{problem}
These connect radius $R$, surface area $S$, and volume $V$ of a sphere:

\begin{enumerate}
\item A sphere has surface area $12\pi$; what is its volume?
\item Find in terms of $R$ the ratio $S/V$; simplify your answer.
\item What is the total surface area of a solid hemisphere of radius $R$?
\item A sphere of radius $R$ is centered at the origin of $xyz$-coordinates. What is the surface area and volume of that portion of it for which $x,y,z\geq 0$?
\end{enumerate}
\end{problem}

\begin{problem}
A sphere is inscribed in a cube, so that it is tangent to all six sides of the cube. What is the ratio (sphere:cube) of their surface areas and volumes?
\end{problem}

\begin{problem}
An upright symmetrical pyramid has a square base with sides of length 10, and a height of 12. What is its volume, and the area of one of its slanted faces?
\end{problem}

\begin{problem}
A right circular cone has radius 2 and height 6. What is the volume of the slice cut off by a plane parallel to the base and distance 2 from it?
\end{problem}

\begin{problem}
A tetrahedron has three edges of lengths 2, 3, and 5 which intersect at right angles. Sketch it and determine its volume.
\end{problem}

\section{Solutions to Geometry Review Problems}

\begin{solution}
a: TODO diagram

Since the triangle is isosceles, the base angles $\alpha$ are equal. 
\[2\alpha+104=180\rightarrow \alpha=38\]
\[x=180-38=142\]
(or $x=104+\alpha=104+38=\boxed{142}$)

b) TODO diagram

Two angles $\alpha$ are equal since lines are parallel. $\alpha=60$ since vertical angles are equal.
\[50+x+60=180\rightarrow \boxed{x=70}\]

c) TODO diagram

Since vertical angles are equal, picture is 

TODO diagram

\[x+20+84=180\rightarrow \boxed{x=76}\]
\end{solution}

\begin{solution}
TODO diagram

The two triangles are similar, so corresponding lines are proportional: \[\frac{OC}{OC'}=\frac{AB}{A'B'}\rightarrow \frac 23=\frac 3{A'B'}\] 

If you prefer, do it in two steps, using similarity of the right triangles: \[\frac{OC}{OC'}=\frac{OA}{OA'}=\frac{AB}{A'B'}\]

In either method, we find $\boxed{A'B'=\frac 92}$.
\end{solution}

\begin{solution}
TODO diagram

Triangles are similar, since bases are parallel lines. Corresponding lines are proportional (c.f. reasoning in previous problem). So \[\frac{x}{100}=\frac{1.2}{2}\rightarrow \boxed{x=60\text{ meters}}\]
\end{solution}

\begin{solution}
a: TODO diagram

The triangles are similar (both have angle $A$ and a right angle) so \[\frac x5=\frac {12}{x+4}\rightarrow x^2+4x=60\rightarrow (x+10)(x-6)=0\] so $\boxed{x=6}$ ($x=-10$ makes no sense.)

b: TODO diagram

$\triangle ABC$ is similar to the small triangle: two equal angles and two right angles. Thus, $\frac x2=\frac{AB}{10}$, and $AB=8$ by the Pythagorean Theorem, so $x=\frac{2\cdot 8}{10}=\boxed{\frac 85}$.
\end{solution}

\begin{solution}
a: TODO diagram

The two angles marked $|alpha$ are equal since both of them equal $90-\beta$. Thus, the left and right triangles are similar, as both have $\alpha$ and a right angle. Thus, $\frac 12=\frac 2x\rightarrow \boxed{x=4}$.

b: TODO diagram

This is the same picture as the above. The bottom and the big triangle are similar (both have angle $A$ and a right angle. Thus, $\frac x1=\frac 13\rightarrow \boxed{x=\frac 13}$.

c: TODO diagram

Triangles are similar, since two angles are equal (alternate interior angles to two parallel lines). Thus, \[\frac 1x=\frac 3{\sqrt{3^2+6^2}}\rightarrow x=\frac{\sqrt{3^2+6^2}}{3}=\frac{3\cdot\sqrt{1+2^2}}3\] so $\boxed{x=\sqrt{5}}$.
\end{solution}

\begin{solution}
a: TODO diagram

Triangles are similar (2 right angles and equal vertical angles). Thus, $\frac 23=\frac hx\rightarrow x=\frac 32 h$. Now, let $A_{top}=\frac{2h}{2}$. Then $A_{bottom}=\frac{3x}{2}=\frac 32\cdot \frac 32 h$ thus $A_{bottom}=\frac 94 A_{top}$.

b: TODO diagram

The two triangles are similar, so corresponding lines are proportional. Thus \[\frac{h_1}{h_2}=\frac 52\rightarrow h_1=\frac 52 h_2\]
\[A_1=\frac{5h_1}{2}\hspace{.5in} A_2=\frac{2h_2}{2}\rightarrow A_1=\frac 52\cdot\frac 52 A_2=\boxed{\frac{25}{4}A}\]
\end{solution}

\begin{solution}
TODO diagram

Two triangles are similar. \[\frac x5=\frac{x+15}{20}\] Crossmultiply: $20x=5x+75\rightarrow x=\boxed{5}$.
\end{solution}

\begin{solution}
TODO diagram

The diagonal is $AD$. We have a right triangle

TODO diagram

so $AD=\sqrt{3}$.

TODO diagram

By same reasoning, $AD=\sqrt{14}$.
\end{solution}

\subsection{Circle problems}
\begin{solution}

a: TODO diagram

$\arck{AB}=180-100=80\rightarrow \angle{ACB}=80\deg$ (central angle is equal to the arc it subtends). Then, $\triangle ACB$ is isosceles so the other two angles are both $50\deg$.

b: TODO diagram

$\angle{A}=100$ (since half the arc it subtends)

$\angle{B}=40$ (same reason)

so $\angle{C}=40$
\end{solution}

\begin{solution}
TODO diagram

Angles $A$ and $B$ are equal, since both subtend the same arc. The two triangles are similar, thus $\frac{h_2}{h_1}=\frac 21$ so $h_2=2h_1$.

TODO diagram

$A_2=\frac{2h_2}{2}, A_1=\frac{1h_1}{2}\rightarrow A_2=h_2=2h_1=4A_1$, so $\boxed{4A}$.
\end{solution}

\begin{solution}
$A$ is a right angle, since it subtends $180\deg$ of arc. Thus the triangles TODO diag and TODO diag are similar (the two angles marked are both complementary to $\angle{A}$, or they subtend equal arcs, by symmetry). Hence, $\frac 2x=\frac x4, x^2=8\rightarrow x=2\sqrt{2}$. Thus, the area of $\triangle ABC$ is $\frac{2\sqrt{2}\cdot 3}{2}=6\sqrt{2}$, and the area of the kite is $\boxed{12\sqrt{2}}$.

TODO diagram

\end{solution}
\begin{solution}
TODO diagram

The angle at $A$ is a right angle (it is inscribed in a semicircle): both smaller triangles are similar to the big one. So, $\frac y1=\frac 4y\rightarrow y=\boxed{2}$ and $\frac x4=\frac 3x\rightarrow x=\boxed{2\sqrt{3}}$.

Check by the Pythagorean Theorem: $x^2+y^2=4^2=(2\sqrt{3})^2+2^2=4^2$.

\end{solution}
\begin{solution}
TODO diagram

$OA\perp PA$ (radius is always perpindicular to tangent) so by Pythagoras, $PO=\sqrt{2^2+1^2}=\sqrt{5}$, so $PB=\boxed{\sqrt{5}-1}$.

\end{solution}
\begin{solution}
TODO diagram

The angle at $A$ is a right angle, since radius is perpindicular to tangent. By Pythagorean Theorem, TODO diag $AB=\sqrt{3^2-1^2}=2\sqrt{2}$ so $CB=\boxed{4\sqrt{2}}$.

\end{solution}
\begin{solution}
TODO diagram

$PB=\sqrt{4^2-2^2}=2\sqrt{3}$.

$A$ and $B$ are right angles (radius perpindicular to tangent). So area of $OPB$ is $\frac{2\cdot 2\sqrt{3}}2=2\sqrt{3}$.

By symmetry, $OBP$ and $OAP$ are congruent, so area $AOBP=\boxed{4\sqrt{3}}$.

\end{solution}
\begin{solution}
TODO diagram

$A$ is a right angle (radius perpendicular to tangent). $OA=OC$ since both are radii; so $OA=\frac 12 OP$ so we have a $30\deg-60\deg-90\deg$ triangle. Thus, $\arck{AC}=90\deg$, so $\arck{ACB}=120\deg$ and $\arck{ADB}=240\deg$.

\end{solution}
\begin{solution}
TODO diagram

The two triangles are similar (the two equal angles marked are alternate interior angles). So $\frac x2=\frac{CB}2\rightarrow x=CB=\sqrt{3^2-2^2}=\boxed{\sqrt{5}}$.
\end{solution}

\subsection{Areas and Volumes}

\begin{solution}
$\frac{360}{72}=5$ so area is $\frac 15$ so area of the circle is $\pi R^2/5$, perimeter is $2R+2\pi R/5$ (see diagram)
\end{solution}

\begin{solution}
TODO diagram

a: TODO diagram

$ABD$ is a 30-60-90 triangle, so $AD=\frac 32\sqrt{3}$ (or use Pythagoras: $AD=\sqrt{3^2-\left(\frac 32\right)^2}=3\sqrt{\frac 34}=\frac 32\sqrt{3}$.)

Thus, area $ABC=\frac 12\cdot 3\cdot\frac 32\sqrt{3}=\boxed{\frac 94\sqrt{3}}$.

b: TODO diagram

$h=2$ (30-60-90 triangle). Thus, area is $\frac 12\cdot 4\cdot 2=\boxed{4}$. (Could also use trigonometry.)
\end{solution}

\begin{solution}
a: TODO diagram

Altitude is $\frac{\sqrt{3}}2$ since we have a 30-60-90 triangle. Thus, the area is $\boxed{3\cdot\frac{\sqrt{3}}2}$.

b: TODO diagram

$h=3$ by Pythagorean Theorem, so area=$3\cdot 16=\boxed{48}$.
\end{solution}

\begin{solution}
TODO diagram

$h=\sqrt{3^2-2^2}=\sqrt{5}$ by Pythagorean Theorem. 

Area is rectangle plus 2 triangles, or $5\sqrt{5}+2\cdot\frac{2\sqrt{5}}2$ (equal to the formula for area of trapezoid)

OR, area is $h$ times the average of the two bases, giving, again, $7\sqrt{5}$.
\end{solution}

\begin{solution}
Let $a$ be the side of square, $r$ the radius of the circle; then $2\pi r=4a\rightarrow r=\frac{2a}{\pi}$. The ratio of the areas is then \[\frac{a^2}{\pi\cdot\frac{4a^2}{\pi^2}}=\boxed{\frac{\pi}{4}}\]
\end{solution}

\begin{solution}
TODO diagram

$A=\pi b^2-\pi a^2=\pi\left(b^2-a^2\right)=\pi(b+a)(b-a)=2\pi\left(\frac{b+a}{2}\right)\left(b-a\right)$. But $2\pi\left(\frac{b+a}{2}\right)$ is the perimeter of the median circle, and $b-a$ is the thickness of the ring, as desired.
\end{solution}

\begin{solution}
TODO diagram

$r=\frac h3$; $V=\pi r^2 h = \pi\frac{h^2}{9}\cdot h=24\pi$. Thus, $h^3=27\cdot 8\rightarrow \boxed{h=6}$, $\boxed{r=2}$.
\end{solution}

\begin{solution}
\begin{enumerate}
\item Surface area is $4\pi r^2=12\pi\rightarrow r=\sqrt{3}$. Volume is $\frac 43\pi r^3=\boxed{4\pi\sqrt{3}}$.
\item \[\frac sv=\frac{4\pi R^2}{\frac 43 \pi R^3}=\boxed{\frac 3R}\]
\item Surface area of hemisphere is half the surface area of the sphere plus the flat side: \[\frac{4\pi R^2}2+\pi R^2=3\pi R^2\]
\item TODO Diag This is $\frac 18$ of both surface area and volume, so surface area is $\boxed{\frac{\pi R^2}2}$ and volume is $\boxed{\frac{\pi R^3}{6}}$.
\end{enumerate}
\end{solution}

\begin{solution}
TODO Diag

Side view: edge of cube and diameter of sphere are of same length. Thus, the area ratio is $\frac{4\pi a^2}{6\cdot(2a)^2}=\boxed{\frac{\pi}{6}}$ and the ratio of the volumes is $\frac{\frac 43 \pi a^3}{(2a)^3}=\boxed{\frac{\pi}{6}}$.
\end{solution}

\begin{solution}
TODO diag side view

volume is $\frac 13$ base ${}\cdot{}$ height, or $\frac 13\cdot 100\cdot 12=\boxed{400}$.

TODO area face diag.

area is $\frac 12\cdot 10\cdot 13=\boxed{65}$.
\end{solution}

\begin{solution}
TODO diag side view

Vol. slice is equal to vol. cone less vol. top cone, or $\frac 13 \pi \cdot 2^2\cdot 6-\frac 13 \pi \left(\frac 43\right)^2\cdot 4=\frac{\pi}3\left(24-\frac{64}{9}\right)=\frac{8\pi}{3}\cdot\frac{19}{9}$. 

(Vol cone is $\frac 13\cdot $ base ${}\cdot{}$ height; by similar triangles, $\frac x2=\frac 46\rightarrow x=\frac 43$.)
\end{solution}

\begin{solution}
TODO diag

Volume is $\frac 13\cdot{}$ base ${}\cdot{}$ height, or $\frac 13\cdot \frac{2\cdot 3}{2}\cdot 5=\boxed{5}$.
\end{solution}


\section{Geometry Diagnostic Test}

\begin{problem}
An irregular pentagon is inscribed in a circle. Four of the sides have the same length, and the other side cuts off an arc on the circle of $100\deg$. What are the angles of the arcs cut off from the other sides?
\end{problem}

\begin{problem}
Two distinct lines are both tangent ot a circle. They meet outside the circle at an angle of $50\deg$. What are the central angles subtended by the two arcs they cut off on the circle? 

TODO diag

$\arck{APB}=?$

$\arck{AQB}=?$
\end{problem}

\section{Geometry Diagnostic Test Solutions}
\begin{sproblem}
An irregular pentagon is inscribed in a circle. Four of the sides have the same length, and the other side cuts off an arc on the circle of $100\deg$. What are the angles of the arcs cut off from the other sides?
\end{sproblem}

\begin{solution}
TODO diagram

$100+4x=360$, so $x=\frac{260}4=\boxed{65\deg}$
\end{solution}

\begin{sproblem}
Two distinct lines are both tangent ot a circle. They meet outside the circle at an angle of $50\deg$. What are the central angles subtended by the two arcs they cut off on the circle? 

TODO diag

$\arck{APB}=?$

$\arck{AQB}=?$
\end{sproblem}

\begin{solution}
Draw in the dotted lines --- they are perpindicular to the tangents.

TODO diag

$\arck{AQB}=360-(90+90+50)=\boxed{130\deg}$

$\arck{APB}=360-130=\boxed{230\deg}$
\end{solution}

