Appendix B — Appendix B: Solutions to Practice

B.1 Chapter 1: The Language of Geometry

B.1.1 Solutions to Practice

  1. Confidence Check (label each statement):
    • “A line segment is a part of a line with two endpoints.” is a Definition.
    • “Through any two distinct points, there is exactly one line.” is a Postulate.
    • “Vertical angles are congruent.” is a Theorem.

B.2 Chapter 2: Star Polygons

B.2.1 Solutions to Practice

  1. Sketch should show a 9-point star using step size 4 (notation \(\{9/4\}\)).

Nine equally spaced dots where every fourth dot is connected.

One valid turning angle label is

\[ \theta=\frac{360^\circ(4)}{9}=160^\circ. \]

  1. \(\{10/2\}\) is disconnected because \(\gcd(10,2)=2>1\). 10 and 2 have common factors, so it is a compound star polygon. Compound star polygons are not connected. You have to lift up your pen/pencil to create it.

  2. \(\{12/3\}\) is disconnected since \(\gcd(12,3)=3\). It splits into 3 identical components. Since 12 and 3 have common factors besides 1, so it is a compound star polygon. Compound star polygons are not connected. You have to lift up your pen/pencil to create it.

  3. Yes. \(\{10/3\}\) and \(\{10/7\}\) are identical because \(\{n/k\}=\{n/(n-k)\}\).

  4. For \(\{7/2\}\),

\[ \theta=\frac{360^\circ(2)}{7}=\frac{720^\circ}{7} \]

\[ \alpha=180^\circ-\theta=180^\circ-\frac{720^\circ}{7}=\frac{540^\circ}{7}\approx77.14^\circ \]

  1. For \(\{11/4\}\),

\[ \theta=\frac{360^\circ(4)}{11}=\frac{1440^\circ}{11}\approx130.91^\circ \]

\[ \alpha=180^\circ-\theta=180^\circ-\frac{1440^\circ}{11}=\frac{540^\circ}{11}\approx49.09^\circ \]

  1. Sketches should show all unique star polygons for \(\{6/k\}\):
Six equally spaced dots where each dot is connected to the next one.
\(\{6/1\}=\{6/5\}\)		 Six equally spaced dots where every second dot is connected.
\(\{6/2\}=\{6/4\}\)		 Six equally spaced dots where every third dot is connected and every dot is used once.
\(\{6/3\}\)

B.3 Chapter 3: Sequences

B.3.1 Solutions to Practice

  1. \(2,6,18,54,\ldots\) is geometric with \(r=3\).
    \(7,12,17,22,\ldots\) is arithmetic with \(d=5\).

  2. For \(4,9,14,19,\ldots\):
    Explicit: \(a_n=4+(n-1)5=5n-1\)
    Recursive: \(a_1=4\), \(a_n=a_{n-1}+5\) for \(n\ge2\).

  3. \(a_{12}=a_1r^{11}=3(2^{11})=6144\).

  4. Arithmetic with \(a_1=10\), \(d=3\), \(a_{15}=52\):

\[ S_{15}=\frac{15}{2}(10+52)=465 \]

  1. Given \(a_1=2\), \(a_n=a_{n-1}+5\):
    Explicit: \(a_n=2+5(n-1)=5n-3\)
    \(a_{10}=5(10)-3=47\).

  2. \(81\left(\frac13\right)^{n-1}=1\) gives \(\left(\frac13\right)^{n-1}=\frac1{81}=\left(\frac13\right)^4\), so \(n=5\).

  3. Differences are \(5,7,9,\ldots\), so next difference is \(11\). Next term is \(24+11=35\).

  4. Geometric with \(a_1=5\), \(r=2\), \(n=8\):

\[ S_8=5\frac{1-2^8}{1-2}=5(255)=1275 \]

B.4 Chapter 4: Figurate Numbers, FIBS, and Sums

B.4.1 Solutions to Practice

\[ T_{12}=\frac{12(13)}{2}=78 \]

  1. First 10 Fibonacci numbers: \(1,1,2,3,5,8,13,21,34,55\).

\[ \sum_{k=1}^{15}k=\frac{15(16)}{2}=120 \]

\[ \sum_{k=1}^{10}k^2=\frac{10(11)(21)}{6}=385 \]

  1. A standard identity is

\[ n^2=T_n+T_{n-1} \]

so each square number is the sum of two consecutive triangular numbers.

\[ P_7=\frac{7(3\cdot7-1)}{2}=\frac{7(20)}{2}=70 \]

\[ 35+36+\cdots+85 \]

There are \(85-35+1=51\) terms, average is \(\frac{35+85}{2}=60\), so sum is \(51(60)=3060\).

  1. Sum of first 20 odd integers is \(20^2=400\).

  2. Sum of first 20 even integers is \(20(21)=420\).

B.5 Chapter 5: Polygons and Polyhedra

B.5.1 Solutions to Practice

  1. Regular pentagon interior angle:

\[ \frac{(5-2)180^\circ}{5}=108^\circ \]

  1. The measure of a central angle equals the measure of its intercepted arc.

  2. Answers depend on the specific diagram used.

  3. For a 14-gon,

\[ \text{Interior sum}=(14-2)180^\circ=2160^\circ \]

If regular, each interior angle is

\[ \frac{2160^\circ}{14}=\frac{1080^\circ}{7}\approx154.29^\circ \]

  1. For a quadrilateral:

\[ (x+10)+(x+20)+(x+30)+(x+40)=360 \]

\[ 4x+100=360 \Rightarrow x=65 \]

Angles: \(75^\circ,85^\circ,95^\circ,105^\circ\).

\[ (n-2)180^\circ=1980^\circ \Rightarrow n-2=11 \Rightarrow n=13 \]

It is a 13-gon (tridecagon).

  1. Using \(F+V=E+2\):

\[ 12+20=E+2 \Rightarrow E=30 \]

B.6 Chapter 6: Tessellations and Fractals

B.6.1 Solutions to Practice

  1. The regular tessellations are by equilateral triangles, squares, and regular hexagons.

  2. A regular pentagon cannot tessellate by itself because its interior angle is \(108^\circ\), and \(360^\circ/108^\circ\) is not an integer.

  3. Triangle-square-hexagon-square gives

\[ 60^\circ+90^\circ+120^\circ+90^\circ=360^\circ \]

so this is a valid vertex arrangement.

    1. Regular
    2. Semi-regular
    3. Neither

  2. With \(N_n=3^n\):
    \(N_4=3^4=81\)
    \(N_6=3^6=729\).

  3. Self-similarity means a pattern contains smaller copies of the same shape at different scales. One example is the Sierpinski triangle.

B.7 Chapter 7: Symmetry, Isometries, and Frieze Patterns

B.7.1 Solutions to Practice

  1. Square: 4 lines. Regular hexagon: 6 lines. Capital letter \(A\): usually 1 vertical line.

  2. For a regular octagon, smallest rotational symmetry angle is

\[ \frac{360^\circ}{8}=45^\circ \]

    1. Translation
    2. Reflection
    3. Rotation
    4. Glide reflection
  2. Apply \((x,y)\to(x-3,y+4)\) to \((5,-2)\):

\[ (5,-2)\to(2,2) \]

  1. Translation plus glide reflection only is frieze type step.

  2. No. A dilation is not an isometry because it changes distances (except the trivial scale factor 1 case).

B.8 Chapter 8: Conic Sections

B.8.1 Solutions to Practice

  1. One correct matching:
    • Circle: plane parallel to the base of the cone
    • Ellipse: plane cuts at an angle and intersects one nappe
    • Parabola: plane parallel to a slant side (generator)
    • Hyperbola: plane cuts both nappes
  2. Cylinder with \(r=4\), \(h=9\):

\[ V=\pi r^2h=\pi(16)(9)=144\pi \]

\[ SA=2\pi rh+2\pi r^2=2\pi(4)(9)+2\pi(16)=104\pi \]

  1. Cone with \(r=5\), \(h=12\):

\[ \ell=\sqrt{r^2+h^2}=\sqrt{25+144}=13 \]

\[ SA=\pi r\ell+\pi r^2=\pi(5)(13)+\pi(25)=90\pi \]

  1. Sphere with \(r=6\):

\[ V=\frac{4}{3}\pi r^3=\frac{4}{3}\pi(216)=288\pi \]

\[ SA=4\pi r^2=4\pi(36)=144\pi \]

  1. True. A circle is a special case of an ellipse (equal semi-axes).

  2. Kepler described planetary orbits as ellipses (with the Sun at a focus). Newton formulated universal gravitation, which explains conic orbits. Hypatia wrote and taught mathematical astronomy and is associated with work/commentary on conic geometry traditions.