Geometry Foundations: Angles, Transformations, and Measurement of Two- and Three-Dimensional Figures

Angles, Rays, and Angle Bisectors

• Nomenclature and Structural Components of Angles:
  • An angle consists of two noncollinear rays sharing a common endpoint called the vertex.
  • The rays forming the angle are designated as the sides of the angle.
  • Angles can be named using the vertex point alone (if unambiguous), three points with the vertex in the middle, or a numerical label assigned to the interior region.

• Angle Identification and Part Analysis:
  • Alternate designations for $\angle 2$: $\angle XZY$ or $\angle YZX$.
  • Vertex of $\angle 3$: Point $Z$.
  • Alternate designations for $\angle 4$: $\angle XYZ$ or $\angle ZYX$.
  • Alternate designations for $\angle YXZ$: $\angle 1$ or $\angle ZXY$.
  • Sides of $\angle 2$: Ray $\vec{ZX}$ and ray $\vec{ZY}$.

• Algebraic Calculations with Bisectors on Opposite Rays (Vertex $T$):
  • Collinear opposite rays $\vec{TP}$ and $\vec{TS}$ form a straight line through vertex $T$.
  • Ray $\vec{TQ}$ bisects $\angle RTP$, establishing the congruence condition $m\angle PTQ = m\angle RTQ$.
  • Given expressions:
  • $m\angle PTQ = (12x + 4)^\circ$
  • $m\angle RTQ = (15x - 5)^\circ$
  • Equating the bisected measures:
  • $12x + 4 = 15x - 5$
  • $3x = 9$
  • $x = 3$
  • Evaluating individual angle measures:
  • $m\angle PTQ = 12(3) + 4 = 40^\circ$
  • $m\angle RTQ = 15(3) - 5 = 40^\circ$
  • Total measure by angle addition:
  • $m\angle RTP = m\angle PTQ + m\angle RTQ = 40^\circ + 40^\circ = 80^\circ$

• Algebraic Calculations with Bisectors on Opposite Rays (Vertex $C$):
  • Opposite rays $\vec{CJ}$ and $\vec{CR}$ define a straight line ($180^\circ$).
  • Ray $\vec{CP}$ bisects $\angle NCW$, implying $m\angle NCP = m\angle PCW$.
  • Ray $\vec{CL}$ bisects $\angle JCN$, implying $m\angle JCL = m\angle LCN$.
  • Calculation 1: Given $m\angle JCL = (6x + 3)^\circ$ and $m\angle LCN = (8x - 13)^\circ$:
  • $6x + 3 = 8x - 13$
  • $2x = 16$
  • $x = 8$
  • $m\angle JCL = 6(8) + 3 = 51^\circ$
  • Calculation 2: Given $m\angle PCW = (4b - 6)^\circ$ and $m\angle WCR = (3b + 2)^\circ$:
  • $m\angle WCR = 26^\circ$
  • Calculation 3: Given $m\angle NCP = (5k - 8)^\circ$ and $m\angle NCW = (8k + 2)^\circ$:
  • Because $\vec{CP}$ bisects $\angle NCW$, $2(m\angle NCP) = m\angle NCW$
  • $2(5k - 8) = 8k + 2$
  • $10k - 16 = 8k + 2$
  • $2k = 18$
  • $k = 9$
  • $m\angle PCW = m\angle NCP = 5(9) - 8 = 37^\circ$
  • Calculation 4: Given $m\angle PCW = 7g$ and $m\angle WCR = 5g + 14$:
  • $m\angle WCR = 49^\circ$
  • Calculation 5: Given $m\angle JCN = (8r - 2)^\circ$ and $m\angle LCN = (3r + 15)^\circ$:
  • Because $\vec{CL}$ bisects $\angle JCN$, $2(m\angle LCN) = m\angle JCN$
  • $2(3r + 15) = 8r - 2$
  • $6r + 30 = 8r - 2$
  • $2r = 32$
  • $r = 16$
  • $m\angle JCL = m\angle LCN = 3(16) + 15 = 63^\circ$
  • Calculation 6: Given $m\angle NCP = (2y + 10)^\circ$ and $m\angle PCW = (4y - 10)^\circ$:
  • $2y + 10 = 4y - 10$
  • $2y = 20$
  • $y = 10$
  • $m\angle PCW = 4(10) - 10 = 30^\circ$

Angle Relationships, Linear Pairs, and Vertical Angles

• Definitions and Geometric Principles:
  • Adjacent Angles: Two coplanar angles that share a common vertex and a common side, but have no interior points in common.
  • Vertical Angles: Two nonadjacent angles formed by two intersecting lines; vertical angles are always congruent.
  • Linear Pair: A pair of adjacent angles whose noncommon sides are opposite rays. The angles in a linear pair are supplementary ($\text{sum} = 180^\circ$).
  • Complementary Angles: Two angles whose measures have a sum of $90^\circ$.
  • Supplementary Angles: Two angles whose measures have a sum of $180^\circ$.

• Angle Pair Identification and Calculation on Intersecting Segments:
  • Adjacent angle example: $\angle ADE$ and $\angle BDE$.
  • Vertical angle example: $\angle DBE$ and $\angle GBC$.
  • Measure calculation for $\angle DEB$: Given adjacent interior angle $\angle FED = 104^\circ$ along straight line segment $FEB$, $m\angle DEB = 180^\circ - 104^\circ = 76^\circ$.

• Algebraic Solving of Unknown Variables in Angle Intersections:

  • Configuration 1: Opposite vertical angles with expressions $(3x + 6)^\circ$ and $54^\circ$:
  • $3x + 6 = 54$
  • $3x = 48$
  • $x = 16$
  • Configuration 2: Opposite vertical angles with expressions $(3x)^\circ$ and $(2x + 44)^\circ$:
  • $3x = 2x + 44$
  • $x = 44$
  • Configuration 3: Opposite vertical angles $(3x - 16)^\circ$ and $(5x - 110)^\circ$, with adjacent supplementary angle $y^\circ$:
  • Equating vertical angles: $5x - 110 = 3x - 16$
  • $2x = 94$
  • $x = 47$
  • Substituting $x$ into the angle measure: $3(47) - 16 = 141 - 16 = 125^\circ$
  • Solving for supplementary variable $y$: $y + 125 = 180$
  • $y = 55$

• Systematic Angle Measure Problems:

  • Problem 1: The measure of the supplement of an angle is four times the measure of the angle.
  • Let the angle be $a$, supplement is $180^\circ - a$
  • $180 - a = 4a$
  • $5a = 180 \implies a = 36^\circ$
  • Supplement $= 180^\circ - 36^\circ = 144^\circ$
  • Problem 2: $\angle M$ and $\angle N$ are complementary. The measure of $\angle M$ is $38^\circ$ less than the measure of $\angle N$.
  • $m\angle M + m\angle N = 90^\circ$
  • $(m\angle N - 38^\circ) + m\angle N = 90^\circ$
  • $2(m\angle N) = 128^\circ \implies m\angle N = 64^\circ$
  • $m\angle M = 64^\circ - 38^\circ = 26^\circ$
  • Problem 3: Two supplementary angles have a difference of $62^\circ$.
  • $x + y = 180^\circ$ and $x - y = 62^\circ$
  • Adding equations: $2x = 242^\circ \implies x = 121^\circ$
  • $y = 180^\circ - 121^\circ = 59^\circ$
  • Problem 4: $\angle S$ and $\angle T$ are complementary. The measure of $\angle S$ is $24^\circ$ more than the measure of $\angle T$.
  • $(m\angle T + 24^\circ) + m\angle T = 90^\circ$
  • $2(m\angle T) = 66^\circ \implies m\angle T = 33^\circ$
  • $m\angle S = 33^\circ + 24^\circ = 57^\circ$
  • Problem 5: The complement of an angle is three times the measure of the angle.
  • $90 - a = 3a$
  • $4a = 90 \implies a = 22.5^\circ$
  • Complement $= 90^\circ - 22.5^\circ = 67.5^\circ$
  • Problem 6: The supplement of an angle is $51^\circ$ less than the angle.
  • $180 - a = a - 51$
  • $2a = 231 \implies a = 115.5^\circ$
  • Supplement $= 180^\circ - 115.5^\circ = 64.5^\circ$
  • Problem 7: $\angle B$ and $\angle C$ are supplementary, with $m\angle B$ being $83^\circ$ greater than $m\angle C$.
  • $(m\angle C + 83^\circ) + m\angle C = 180^\circ$
  • $2(m\angle C) = 97^\circ \implies m\angle C = 48.5^\circ$
  • $m\angle B = 48.5^\circ + 83^\circ = 131.5^\circ$
  • Problem 8: The complement of an angle is five times the measure of the angle.
  • $90 - a = 5a$
  • $6a = 90 \implies a = 15^\circ$
  • Complement $= 90^\circ - 15^\circ = 75^\circ$

• Proportional Circular Arc Motion Application:

  • A turtle walks along the perimeter of a circular fountain starting at point $A$.
  • Walking from point $A$ to point $B$ spans a central angle of $24^\circ$ and requires $22\,\text{seconds}$.
  • Angular walking rate:
  • $\text{Rate} = \frac{24^\circ}{22\,\text{s}} = \frac{12}{11}\,^\circ/\text{s}$
  • Point $C$ is positioned exactly half-way around the fountain perimeter from point $A$, representing a central angle displacement of $180^\circ$.
  • Remaining angular distance from point $B$ to point $C$:
  • $\Delta \theta = 180^\circ - 24^\circ = 156^\circ$
  • Additional time required to reach point $C$:
  • $t = \frac{156^\circ}{\frac{12}{11}\,^\circ/\text{s}} = 156 \times \frac{11}{12} = 13 \times 11 = 143\,\text{s}$

• Perpendicular Lines and Angle Relationships:

  • Intersecting lines $\overleftrightarrow{AB} \perp \overleftrightarrow{CD}$ at point $E$:
  • $m\angle AED = (6x - 24)^\circ = 90^\circ \implies 6x = 114 \implies x = 19$
  • $m\angle CEB = (4y + 32)^\circ = 90^\circ \implies 4y = 58 \implies y = 14.5$
  • Perpendicular rays $\vec{RS} \perp \vec{RT}$ with interior point $V$:
  • $m\angle SRV + m\angle VRT = 90^\circ$
  • $(9j - 44) + (6j + 14) = 90$
  • $15j - 30 = 90 \implies 15j = 120 \implies j = 8$
  • $m\angle SRV = 9(8) - 44 = 72 - 44 = 28^\circ$
  • $m\angle VRT = 6(8) + 14 = 48 + 14 = 62^\circ$
  • Perpendicular rays $\vec{QR} \perp \vec{QS}$ with interior point $T$:
  • $m\angle RQT + m\angle SQT = 90^\circ$
  • $(7w + 12) + (9w - 18) = 90$
  • $16w - 6 = 90 \implies 16w = 96 \implies w = 6$
  • $m\angle RQT = 7(6) + 12 = 42 + 12 = 54^\circ$
  • $m\angle SQT = 9(6) - 18 = 54 - 18 = 36^\circ$

• Angle Relationships with Perpendicular Rays and Straight Lines:
  • In the figure, straight line $\overleftrightarrow{NK}$ has vertex $L$, with $\vec{LH} \perp \vec{LJ}$, so $m\angle HLJ = 90^\circ$.
  • Sub-problem A: Given $m\angle HLG = (8p - 17)^\circ$ and $m\angle GLJ = (5p + 3)^\circ$:
  • $(8p - 17) + (5p + 3) = 90$
  • $13p - 14 = 90 \implies 13p = 104 \implies p = 8$
  • $m\angle HLG = 8(8) - 17 = 64 - 17 = 47^\circ$
  • $m\angle GLJ = 5(8) + 3 = 40 + 3 = 43^\circ$
  • Sub-problem B: Given straight angle $m\angle NLK = 180^\circ$ and right angle $m\angle HLJ = 90^\circ$, the exterior angles sum to $90^\circ$:
  • $m\angle NLH + m\angle JLK = 180^\circ - 90^\circ = 90^\circ$
  • $(3s + 5) + (6s - 23) = 90$
  • $9s - 18 = 90 \implies 9s = 108 \implies s = 12$
  • $m\angle NLH = 3(12) + 5 = 36 + 5 = 41^\circ$
  • $m\angle JLK = 6(12) - 23 = 72 - 23 = 49^\circ$

• Geometric Deductions from Diagram Configurations:
  • Statement: $\angle 4$ and $\angle 7$ are supplementary.
  • Determination: Yes; they share a common side and are formed by two intersecting straight lines, establishing them as a linear pair.
  • Statement: $\angle 3$ and $\angle 10$ are complementary.
  • Determination: Yes; because $\angle 12$ is explicitly designated as a right angle ($90^\circ$), the adjacent angles completing the straight line must sum to $90^\circ$.
  • Statement: $\angle 4$ and $\angle 5$ form a linear pair.
  • Determination: No; the angles do not share a common side.
  • Statement: $\angle 2$ and $\angle 10$ are vertical angles.
  • Determination: No; these angles are not formed by the intersection of exactly two straight lines.
  • Statement: $\angle 1$ and $\angle 2$ form a linear pair.
  • Determination: Yes; they share a common side and their noncommon sides form a straight line.
  • Statement: $\angle 7$ and $\angle 9$ are vertical angles.
  • Determination: Yes; they are nonadjacent angles formed by two straight intersecting lines.

Two-Dimensional Geometric Figures, Perimeters, and Areas

• Formula Reference for Planar Figures:
  • Triangle:
  • $\text{Perimeter} = a + b + c$
  • $\text{Area} = \frac{1}{2} b h$
  • Rectangle / Parallelogram:
  • $\text{Perimeter} = 2l + 2w$
  • $\text{Area} = l \times w$
  • Circle:
  • $\text{Circumference} = 2\pi r = \pi d$
  • $\text{Area} = \pi r^2$
  • Distance formula on a coordinate plane:
  • $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• Perimeter/Circumference and Area on a Coordinate Grid ($1\,\text{unit} = 1\,\text{cm}$):

  • Grid Item 1: Circle with center $A(5, 3)$ and point on circumference $B(2, 1)$.
  • Radius $r = \sqrt{(5 - 2)^2 + (3 - 1)^2} = \sqrt{3^2 + 2^2} = \sqrt{13} \approx 3.6056\,\text{cm}$
  • $\text{Circumference} = 2\pi \sqrt{13} \approx 22.7\,\text{cm}$
  • $\text{Area} = \pi (\sqrt{13})^2 = 13\pi \approx 40.8\,\text{cm}^2$
  • Grid Item 2: Triangle with vertices $A(2, 1)$, $B(8, 6)$, and $C(10, 1)$.
  • Horizontal base length $b = 10 - 2 = 8\,\text{cm}$
  • Height $h = 6 - 1 = 5\,\text{cm}$
  • Side lengths: $AB = \sqrt{6^2 + 5^2} = \sqrt{61} \approx 7.8102\,\text{cm}$; $BC = \sqrt{2^2 + 5^2} = \sqrt{29} \approx 5.3852\,\text{cm}$
  • $\text{Perimeter} = 8 + \sqrt{61} + \sqrt{29} \approx 21.2\,\text{cm}$
  • $\text{Area} = \frac{1}{2}(8)(5) = 20\,\text{cm}^2$
  • Grid Item 3: Square with vertices $A(-5, 5)$, $B(0, 3)$, $C(-2, -2)$, and $D(-7, 0)$.
  • Side length $s = \sqrt{(0 - (-5))^2 + (3 - 5)^2} = \sqrt{5^2 + (-2)^2} = \sqrt{29} \approx 5.3852\,\text{cm}$
  • $\text{Perimeter} = 4\sqrt{29} \approx 21.5\,\text{cm}$
  • $\text{Area} = (\sqrt{29})^2 = 29\,\text{cm}^2$
  • Grid Item 4: Rectangle with vertices $A(2, 6)$, $B(5, 3)$, $C(1, -1)$, and $D(-2, 2)$.
  • Side length $AB = \sqrt{3^2 + (-3)^2} = \sqrt{18} \approx 4.2426\,\text{cm}$
  • Side length $BC = \sqrt{(-4)^2 + (-4)^2} = \sqrt{32} \approx 5.6569\,\text{cm}$
  • $\text{Perimeter} = 2(\sqrt{18} + \sqrt{32}) \approx 19.8\,\text{cm}$
  • $\text{Area} = \sqrt{18} \times \sqrt{32} = \sqrt{576} = 24.0\,\text{cm}^2$
  • Grid Item 5: Triangle with vertices $A(-6, 6)$, $B(2, 3)$, and $C(-3, -2)$.
  • Side lengths: $AB = \sqrt{8^2 + (-3)^2} = \sqrt{73} \approx 8.5440\,\text{cm}$; $BC = \sqrt{(-5)^2 + (-5)^2} = \sqrt{50} \approx 7.0711\,\text{cm}$; $AC = \sqrt{3^2 + (-8)^2} = \sqrt{73} \approx 8.5440\,\text{cm}$
  • $\text{Perimeter} = 2\sqrt{73} + \sqrt{50} \approx 24.2\,\text{cm}$
  • $\text{Area} = 27.5\,\text{cm}^2$
  • Grid Item 6: Circle with center $A(2, 2)$ and point $B(5, 5)$.
  • Radius $r = \sqrt{(5 - 2)^2 + (5 - 2)^2} = \sqrt{3^2 + 3^2} = \sqrt{18} \approx 4.2426\,\text{cm}$
  • $\text{Circumference} = 2\pi \sqrt{18} \approx 26.7\,\text{cm}$
  • $\text{Area} = \pi (\sqrt{18})^2 = 18\pi \approx 56.5\,\text{cm}^2$

• Physical Object Models (Two-Dimensional Models):

  • Triangular sandwich slice:
  • Side lengths: $12\,\text{cm}$, $10.8\,\text{cm}$, and base split into segments $8.4\,\text{cm}$ and $6.6\,\text{cm}$ ($\text{total base} = 15.0\,\text{cm}$
  • Perpendicular altitude to base: $8.6\,\text{cm}$
  • $\text{Perimeter} = 12 + 10.8 + 15.0 = 37.8\,\text{cm}$
  • $\text{Area} = \frac{1}{2}(15.0)(8.6) = 64.5\,\text{cm}^2$
  • Circular plate/coaster:
  • Diameter $d = 7.2\,\text{cm} \implies \text{radius } r = 3.6\,\text{cm}$
  • $\text{Circumference} = 7.2\pi \approx 22.6\,\text{cm}$
  • $\text{Area} = \pi (3.6)^2 = 12.96\pi \approx 40.7\,\text{cm}^2$
  • Tablet computer:
  • Width $= 20.2\,\text{in.}$, Height $= 13.4\,\text{in.}$
  • $\text{Perimeter} = 2(20.2 + 13.4) = 67.2\,\text{in.}$
  • $\text{Area} = 20.2 \times 13.4 = 270.7\,\text{in.}^2$
  • Circular alarm clock face:
  • Diameter $= 4.5\,\text{in.} \implies r = 2.25\,\text{in.}$
  • $\text{Circumference} = 4.5\pi \approx 14.1\,\text{in.}$
  • $\text{Area} = \pi (2.25)^2 \approx 15.9\,\text{in.}^2$
  • Rectangular table top:
  • Length $= 3.6\,\text{ft}$, Width $= 2.8\,\text{ft}$
  • $\text{Perimeter} = 2(3.6 + 2.8) = 12.8\,\text{ft}$
  • $\text{Area} = 3.6 \times 2.8 = 10.1\,\text{ft}^2$
  • Triangular tent profile:
  • Base $= 6\,\text{ft } 4\,\text{in.} = 6.\bar{3}\,\text{ft}$
  • Equal slant sides $= 7\,\text{ft } 10\,\text{in.} = 7.8\bar{3}\,\text{ft}$
  • Height $= 7\,\text{ft } 2\,\text{in.} = 7.1\bar{6}\,\text{ft}$
  • $\text{Perimeter} = 7.8\bar{3} + 7.8\bar{3} + 6.\bar{3} = 22\,\text{ft}$
  • $\text{Area} = \frac{1}{2}(6.\bar{3})(7.1\bar{6}) \approx 22.7\,\text{ft}^2$

• Real-World Composite Boundary and Area Problems:
  • Backyard Landscaping Project:
  • Yard boundary: Rectangular region of width $32\,\text{ft} + 8\,\text{ft} = 40\,\text{ft}$ and depth $18\,\text{ft}$.
  • Gate installations: Two $3\,\text{ft}$ gaps on either side of the house along the front edge.
  • Total fencing required (including sides, back, and two front gates):
    • $\text{Fencing} = 18 + 40 + 18 + 3 + 3 = 82\,\text{ft}$
  • Triangular garden: Right triangle at back-right corner with perpendicular legs of $8\,\text{ft}$ and $13\,\text{ft}$.
    • $\text{Garden Area} = \frac{1}{2}(8)(13) = 52\,\text{ft}^2$
  • Grass seed area (remaining area of yard):
    • $\text{Total Yard Area} = 40 \times 18 = 720\,\text{ft}^2$
    • $\text{Grass Area} = 720 - 52 = 668\,\text{ft}^2$
  • Quilting Pattern Calculation:
  • Quilt dimensions: $7\,\text{ft } 4\,\text{in.}$ by $6\,\text{ft } 5\,\text{in.}$ ($88\,\text{in.}$ by $77\,\text{in.}$).
  • Modular square size: $11\,\text{in.}$ by $11\,\text{in.}$; each contains a circle of diameter $6\,\text{in.}$ ($\text{radius } r = 3\,\text{in.}$).
  • Total squares needed:
    • Squares along length: $\frac{88}{11} = 8$
    • Squares along width: $\frac{77}{11} = 7$
    • $\text{Total Count} = 8 \times 7 = 56\text{ squares}$
  • Area of individual square: $11 \times 11 = 121\,\text{in.}^2$
  • Total area of all 56 squares: $56 \times 121 = 6776\,\text{in.}^2$
  • Area of individual circular patch: $\pi (3)^2 = 9\pi \approx 28.3\,\text{in.}^2$
  • Total area of all 56 circles: $56 \times 9\pi = 504\pi \approx 1583.4\,\text{in.}^2$
  • Total length of outer quilt border:
    • $\text{Border Perimeter} = 2(88 + 77) = 2(165) = 330\,\text{in.}$
  • Art Classroom Table Coverings:
  • Rectangular tables: $6\,\text{ft}$ by $2\,\text{ft}$.
    • $\text{Paper Area per Rectangular Table} = 6 \times 2 = 12\,\text{ft}^2$
  • Circular tables: Diameter $2\,\text{ft } 6\,\text{in.} = 2.5\,\text{ft} \implies r = 1.25\,\text{ft}$.
    • $\text{Paper Area per Circular Table} = \pi (1.25)^2 \approx 4.9\,\text{ft}^2$
  • Total covering for 12 rectangular tables and 6 circular tables:
    • $\text{Total Area} = 12(12) + 6(\pi \times 1.25^2) = 144 + 9.375\pi \approx 144 + 29.452 = 173.5\,\text{ft}^2$
  • Semicircular and Square Composite Window:
  • Bottom section: Square of side length $4\,\text{ft } 6\,\text{in.} = 54\,\text{in.}$, divided into a $3 \times 3$ grid of 9 identical square panes.
  • Top section: Semicircular pane with diameter $54\,\text{in.}$ ($\text{radius } r = 27\,\text{in.}$).
  • Area of semicircular pane:
    • $\text{Area} = \frac{1}{2}\pi (27)^2 = 364.5\pi \approx 1145.1\,\text{in.}^2$
  • Area of each individual square pane (side $= \frac{54}{3} = 18\,\text{in.}$):
    • $\text{Area} = 18 \times 18 = 324\,\text{in.}^2$
  • Total perimeter of the composite window:
    • Bottom edge: $54\,\text{in.}$
    • Left and right vertical edges: $54 + 54 = 108\,\text{in.}$
    • Curved top edge: $\pi r = 27\pi \approx 84.823\,\text{in.}$
    • $\text{Total Perimeter} = 54 + 108 + 27\pi \approx 246.8\,\text{in.}$

Rigid Transformations in the Coordinate Plane

• Fundamental Isometric Transformations:
  • Rigid motions preserve distance, angle measure, and area between pre-image and image.
  • Reflection: An isometry that maps each point across a specified line of reflection acting as a perpendicular bisector to the segment connecting the pre-image and image points.
  • Translation: An isometry that shifts every point of a figure by the same distance in the specified direction of a vector $\langle a, b \rangle$.
  • Rotation: An isometry that turns every point of a figure through a specified angle and direction about a fixed center of rotation.

• Identification of Rigid Motions in Real-World Contexts:
  • Turning bicycle wheel: Rotation
  • Train advancing straight along tracks: Translation
  • Moon reflecting on a calm water surface: Reflection
  • Hand rotating a combination lock dial: Rotation

• Identification of Geometric Transformations on Graphs:

  • Graph 5: Green triangle reflected across the $y$-axis to form blue triangle: Reflection
  • Graph 6: Blue polygon shifted vertically down into green polygon: Translation
  • Graph 7: Trapezoid inverted across the horizontal axis: Reflection or rotation
  • Graph 8: Green parallelogram mapped to blue parallelogram by turning about the origin: Rotation

• Coordinate Mapping Rules:

  • Reflection in the $x$-axis: $(x, y) \rightarrow (x, -y)$
  • Reflection in the $y$-axis: $(x, y) \rightarrow (-x, y)$
  • Translation along vector $\langle a, b \rangle$: $(x, y) \rightarrow (x + a, y + b)$
  • Counterclockwise rotation of $90^\circ$ about the origin: $(x, y) \rightarrow (-y, x)$
  • Rotation of $180^\circ$ about the origin: $(x, y) \rightarrow (-x, -y)$
  • Counterclockwise rotation of $270^\circ$ about the origin: $(x, y) \rightarrow (y, -x)$

• Transformations of Triangle $JKL$ with Vertices $J(-1, 5)$, $K(1, 2)$, $L(-3, -1)$:

  • Reflection in $x$-axis:
  • $J'(-1, -5)$, $K'(1, -2)$, $L'(-3, 1)$
  • Reflection in $y$-axis:
  • $J'(1, 5)$, $K'(-1, 2)$, $L'(3, -1)$
  • Translation along vector $\langle 5, -4 \rangle$:
  • $J'(-1 + 5, 5 - 4) = J'(4, 1)$
  • $K'(1 + 5, 2 - 4) = K'(6, -2)$
  • $L'(-3 + 5, -1 - 4) = L'(2, -5)$
  • Translation along vector $\langle -2, -3 \rangle$:
  • $J'(-1 - 2, 5 - 3) = J'(-3, 2)$
  • $K'(1 - 2, 2 - 3) = K'(-1, -1)$
  • $L'(-3 - 2, -1 - 3) = L'(-5, -4)$
  • Rotation $90^\circ$ counterclockwise about origin:
  • $J'(-5, -1)$, $K'(-2, 1)$, $L'(1, -3)$
  • Rotation $180^\circ$ about origin:
  • $J'(1, -5)$, $K'(-1, -2)$, $L'(3, 1)$

• Transformations of Triangle $JKL$ with Vertices $J(2, -5)$, $K(-1, -3)$, $L(-4, 5)$:

  • Reflection in $x$-axis:
  • $J'(2, 5)$, $K'(-1, 3)$, $L'(-4, -5)$
  • Reflection in $y$-axis:
  • $J'(-2, -5)$, $K'(1, -3)$, $L'(4, 5)$
  • Translation along vector $\langle -3, 1 \rangle$:
  • $J'(2 - 3, -5 + 1) = J'(-1, -4)$
  • $K'(-1 - 3, -3 + 1) = K'(-4, -2)$
  • $L'(-4 - 3, 5 + 1) = L'(-7, 6)$
  • Rotation $180^\circ$ about origin:
  • $J'(-2, 5)$, $K'(1, 3)$, $L'(4, -5)$
  • Rotation $270^\circ$ counterclockwise about origin:
  • Applying $(x, y) \rightarrow (y, -x)$
  • $J'(-5, -2)$, $K'(-3, 1)$, $L'(5, 4)$

• Stage Vector Shift Problem:

  • Pre-image positions of actors: $(-4, 8)$, $(2, 2)$, and $(-6, -4)$.
  • Target image positions of actors: $(0, 6)$, $(6, 0)$, and $(-2, -6)$.
  • Vector component evaluation:
  • Horizontal shift $\Delta x = 0 - (-4) = 6 - 2 = -2 - (-6) = 4$
  • Vertical shift $\Delta y = 6 - 8 = 0 - 2 = -6 - (-4) = -2$
  • Resulting transformation description: Translation along vector $\langle 4, -2 \rangle$.

Three-Dimensional Figures, Polyhedra, Surface Area, and Volume

• Classification and Terminology of Three-Dimensional Solids:
  • Polyhedron: A closed three-dimensional figure bounded by flat polygonal surfaces called faces.
  • Edges: The line segments where two polygonal faces intersect.
  • Vertices: The points where three or more edges intersect.
  • Prism: A polyhedron with two congruent parallel polygonal bases connected by rectangular lateral faces.
  • Pyramid: A polyhedron with one polygonal base and triangular lateral faces meeting at a single apex vertex.
  • Non-polyhedra: Solids featuring curved surfaces, including cylinders, cones, and spheres.

• Identification of Solids and Solid Anatomy:

  • Solid 1: Cylinder; not a polyhedron.
  • Solid 2: Triangular prism; polyhedron.
  • Triangular bases: $\triangle PQR$ and $\triangle STU$
  • Rectangular faces: $\square STQP$, $\square UTQR$, and $\square USPR$
  • Edges: $\overline{RP}$, $\overline{PQ}$, $\overline{QR}$, $\overline{ST}$, $\overline{TU}$, $\overline{US}$, $\overline{RU}$, $\overline{PS}$, and $\overline{QT}$
  • Vertices: Points $P$, $Q$, $R$, $S$, $T$, and $U$
  • Solid 3: Sphere with center $W$; not a polyhedron.
  • Physical Object 4: Square pyramid; polyhedron.
  • Physical Object 5: Volcano cone; not a polyhedron.
  • Physical Object 6: Game die (cube); polyhedron.

• Surface Area and Volume Formulas:

  • Right Cone:
  • $\text{Surface Area } T = \pi r^2 + \pi r l$
  • $\text{Volume } V = \frac{1}{3} \pi r^2 h$
  • Sphere:
  • $\text{Surface Area } T = 4\pi r^2$
  • $\text{Volume } V = \frac{4}{3} \pi r^3$
  • Right Cylinder:
  • $\text{Surface Area } T = 2\pi r^2 + 2\pi r h$
  • $\text{Volume } V = \pi r^2 h$
  • Regular Pyramid:
  • $\text{Surface Area } T = B + \frac{1}{2} P l$
  • $\text{Volume } V = \frac{1}{3} B h$
  • Rectangular Prism:
  • $\text{Surface Area } T = 2(lw + lh + wh)$
  • $\text{Volume } V = l \times w \times h$
  • Triangular Prism:
  • $\text{Surface Area } T = 2B + P h$
  • $\text{Volume } V = B h$

• Surface Area and Volume Computational Problems:

  • Problem 7: Right cone with base diameter $8\,\text{ft}$ ($r = 4\,\text{ft}$), height $h = 12\,\text{ft}$, and slant height $l = 12.6\,\text{ft}$.
  • $T = \pi (4)^2 + \pi (4)(12.6) = 16\pi + 50.4\pi = 66.4\pi \approx 208.6\,\text{ft}^2$
  • $V = \frac{1}{3}\pi (4)^2 (12) = 64\pi \approx 201.1\,\text{ft}^3$
  • Problem 8: Sphere with radius $r = 5.1\,\text{m}$.
  • $T = 4\pi (5.1)^2 = 104.04\pi \approx 326.9\,\text{m}^2$
  • $V = \frac{4}{3}\pi (5.1)^3 = 176.868\pi \approx 555.6\,\text{m}^3$
  • Problem 9: Cylinder with radius $r = 7.5\,\text{in.}$ and height $h = 15\,\text{in.}$.
  • $T = 2\pi (7.5)^2 + 2\pi (7.5)(15) = 112.5\pi + 225\pi = 337.5\pi \approx 1060.3\,\text{in.}^2$
  • $V = \pi (7.5)^2 (15) = 843.75\pi \approx 2650.7\,\text{in.}^3$
  • Problem 10: Inverted square pyramid with base edges $18\,\text{ft}$, height $h = 13.2\,\text{ft}$, and slant height $l = 16\,\text{ft}$.
  • Base area $B = 18 \times 18 = 324\,\text{ft}^2$
  • Base perimeter $P = 4(18) = 72\,\text{ft}$
  • $T = 324 + \frac{1}{2}(72)(16) = 324 + 576 = 900\,\text{ft}^2$
  • $V = \frac{1}{3}(324)(13.2) = 108(13.2) = 1425.6\,\text{ft}^3$
  • Problem 11: Rectangular prism with length $7\,\text{yd}$, width $3\,\text{yd}$, and height $4\,\text{yd}$.
  • $T = 2(7 \times 3 + 7 \times 4 + 3 \times 4) = 2(21 + 28 + 12) = 2(61) = 122\,\text{yd}^2$
  • $V = 7 \times 3 \times 4 = 84\,\text{yd}^3$
  • Problem 12: Right triangular prism with base legs $8.3\,\text{cm}$ and $6.2\,\text{cm}$, hypotenuse $10.4\,\text{cm}$, and prism depth $2.8\,\text{cm}$.
  • Base triangular area $B = \frac{1}{2}(8.3)(6.2) = 25.73\,\text{cm}^2$
  • Base perimeter $P = 8.3 + 6.2 + 10.4 = 24.9\,\text{cm}$
  • $T = 2(25.73) + 24.9(2.8) = 51.46 + 69.72 = 121.18 \approx 121.2\,\text{cm}^2$
  • $V = 25.73 \times 2.8 = 72.044 \approx 72.0\,\text{cm}^3$

• Applied Engineering and Scientific 3D Calculations:

  • Stainless Steel Cylinder Water Bottle:
  • Design: Height $= 7\,\text{in.}$, radius $= 1.5\,\text{in.}$. Steel comprises one circular base and the curved lateral surface; top lid is plastic.
  • Steel surface area needed:
    • $SA = \pi r^2 + 2\pi r h = \pi (1.5)^2 + 2\pi (1.5)(7) = 2.25\pi + 21\pi = 23.25\pi \approx 73.0\,\text{in.}^2$
  • Volume capacity of liquid:
    • $V = \pi (1.5)^2 (7) = 15.75\pi \approx 49.5\,\text{in.}^3$
  • Earth and Atmospheric Shell:
  • Mean Earth diameter $= 8000\,\text{mi} \implies \text{radius } R_1 = 4000\,\text{mi}$.
  • Atmospheric shell thickness $= 60\,\text{mi} \implies \text{outer radius } R_2 = 4060\,\text{mi}$.
  • Earth surface area:
    • $T_{\text{Earth}} = 4\pi (4000)^2 = 64,000,000\pi \approx 201,061,930\,\text{mi}^2$
  • Volume of Earth's atmosphere (spherical shell difference):
    • $V_{\text{atm}} = \frac{4}{3}\pi (R_2^3 - R_1^3) = \frac{4}{3}\pi (4060^3 - 4000^3)$
    • $4060^3 = 66,923,416,000$
    • $4000^3 = 64,000,000,000$
    • $\Delta V = 2,923,416,000$
    • $V_{\text{atm}} = \frac{4}{3}\pi (2,923,416,000) = 3,897,888,000\pi \approx 12,245,576,305\,\text{mi}^3$
  • Ferris Wheel Gondola Seat Capacity:
  • Vertical diameter $d = 48\,\text{m}$.
  • Circumference $C = 48\pi \approx 150.796\,\text{m}$.
  • Gondolas placed every $5\,\text{m}$ along perimeter:
    • $\text{Maximum Seats} = \left\lfloor \frac{150.796}{5} \right\rfloor = \lfloor 30.159 \rfloor = 30\text{ seats}$
  • Parking Lot Gravel Bed Layer:
  • Gravel depth $= 6\,\text{in.} = 0.5\,\text{ft}$.
  • Unpaved rectangle: Width $= 100\,\text{ft}$, Length $= 200\,\text{ft}$.
  • Volume of gravel:
    • $V = 100 \times 200 \times 0.5 = 10,000\,\text{ft}^3$
  • Dump truck capacity: Bed is $8\,\text{ft}$ wide, $12\,\text{ft}$ long, and $5\,\text{ft}$ deep ($\text{total volume} = 480\,\text{ft}^3$).
  • Given truck fractional loading condition: Delivering $10,000\,\text{ft}^3$ requires purchasing $32\text{ trucks}$.