Math Coverage Grade 8 Advanced EOT 3: Modules 8, 9, 10, and 11

Module 8: Transformations

M8L1: Translations

A translation is a type of transformation that slides a figure from one position to another without turning it. The resulting shape is the image, and the original shape is the preimage.

Coordinate Notation

You can use coordinate notation to describe a translation as follows: (x,y)→(x+a,y+b)(x, y) \rightarrow (x + a, y + b)

  • Horizontal Shift (x+ax + a): Add aa to the x-coordinate. If aa is positive, move to the right; if negative, move to the left.

  • Vertical Shift (y+by + b): Add bb to the y-coordinate. If bb is positive, move up; if negative, move down.

M8L2: Reflections

A reflection is a mirror image of the original figure. It occurs across a line called the line of reflection.

Reflection Rules
  • Across the x-axis: Keep the x-coordinate the same and use the opposite of the y-coordinate. (x,y)→(x,−y)(x, y) \rightarrow (x, -y).

  • Across the y-axis: Keep the y-coordinate the same and use the opposite of the x-coordinate. (x,y)→(−x,y)(x, y) \rightarrow (-x, y).

Specific Reflection Examples and Practice
  • Reflection across the line x=2x = 2: See Problem 11 for ∆CDE∆CDE.

  • Reflection across the line y=−1y = -1: See Problem 12 for polygon FGHIFGHI.

  • Problem 13: Triangle TUVTUV has coordinates T(0,3)T(0, 3), U(−3,0)U(-3, 0), and V(−4,4)V(-4, 4). Reflected across the y-axis, the notation is (x,y)→(−x,y)(x, y) \rightarrow (-x, y). The image coordinates are T′(0,3)T'(0, 3), U′(3,0)U'(3, 0), and V′(4,4)V'(4, 4).

  • Problem 14: ∆LMN∆LMN maps to ∆L′M′N′∆L'M'N'. Given M(−4,1)→M′(−4,−1)M(-4, 1) \rightarrow M'(-4, -1) and N(−1,3)→N′(−1,−3)N(-1, 3) \rightarrow N'(-1, -3), the transformation is a reflection across the x-axis.

M8L3: Rotations

A rotation is a transformation where a figure is turned about a fixed point. Rotations can be clockwise or counterclockwise.

Clockwise Rotation Rules
  • 90∘90^{\circ} Clockwise: Swap xx and yy, then change the sign of the new y-coordinate. (x,y)→(y,−x)(x, y) \rightarrow (y, -x).

  • 180∘180^{\circ} Clockwise: Change the sign of both coordinates. (x,y)→(−x,−y)(x, y) \rightarrow (-x, -y).

  • 270∘270^{\circ} Clockwise (or 90∘90^{\circ} counterclockwise): Swap xx and yy, then change the sign of the new x-coordinate. (x,y)→(−y,x)(x, y) \rightarrow (-y, x).

M8L4: Dilations

A dilation is a transformation that enlarges or reduces a figure relative to a center point by a scale factor (kk).

Coordinate Notation

(x,y)→(kx,ky)(x, y) \rightarrow (kx, ky) You must multiply both the x and y coordinates by the scale factor kk.

  • If k>1k > 1, the figure gets bigger (enlargement).

  • If k<1k < 1, the figure gets smaller (reduction).

Dilation Examples
  • Problem 23: Trapezoid RAINRAIN with vertices R(−2,1)R(-2, 1), A(1,1)A(1, 1), I(0,−1)I(0, -1), and N(−1,−1)N(-1, -1) dilated by k=2k = 2. Coordinates become R′(−4,2)R'(-4, 2), A′(2,2)A'(2, 2), I′(0,−2)I'(0, -2), and N′(−2,−2)N'(-2, -2).

  • Problem 25: Triangle JKLJKL with vertices J(−4,−1)J(-4, -1), K(0,4)K(0, 4), and L(−4,−2)L(-4, -2) dilated by k=0.5k = 0.5. Coordinates become J′(−2,−0.5)J'(-2, -0.5), K′(0,2)K'(0, 2), and L′(−2,−1)L'(-2, -1).

  • Problem 26: Rectangle PQRSPQRS with vertices P(−3,3)P(-3, 3), Q(6,3)Q(6, 3), R(6,−3)R(6, -3), and S(−3,−3)S(-3, -3) dilated by k=13k = \frac{1}{3}. Coordinates become P′(−1,1)P'(-1, 1), Q′(2,1)Q'(2, 1), R′(2,−1)R'(2, -1), and S′(−1,−1)S'(-1, -1).

  • Problem 28: Keisha used a photo measuring 4 inches×6 inches4\,inches × 6\,inches to make a copy measuring 8 inches×12 inches8\,inches × 12\,inches. The scale factor k=84=2k = \frac{8}{4} = 2.

Module 9: Congruence and Similarity

M9L1 & M9L2: Congruence Transformations and Corresponding Parts

  • Congruence Transformation: A transformation (like translation, reflection, or rotation) that keeps the figure the same shape and same size.

  • Corresponding Parts:

    • Sides: Sides with the same number of tick marks are congruent.

    • Angles: Angles with the same number of arc marks are congruent.

  • Problem 39 (Quilt Design): ∆RST≅∆RWX∆RST \cong ∆RWX. If m∠WXR=62∘m\angle WXR = 62^{\circ}, then its corresponding angle m∠STR=62∘m\angle STR = 62^{\circ}.

  • Problem 40 (Baseball Diamond): Given ∆BEA≅∆ARB∆BEA \cong ∆ARB. If the length of BE=90 ftBE = 90\,ft, then the length of the corresponding side AR=90 ftAR = 90\,ft.

M9L3 & M9L4: Similarity and Transformations

  • Similarity: Two figures are similar if they have the same shape but not necessarily the same size. Dilation is the primary transformation that creates similarity.

  • Properties of Similar Polygons:

    1. Corresponding angles are congruent (m∠A=m∠Xm\angle A = m\angle X, m∠B=m∠Ym\angle B = m\angle Y, etc.).

    2. Lengths of corresponding sides are proportional (ABXY=BCYZ=ACXZ\frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}).

  • Angle-Angle (AA) Similarity Theorem: If two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar.

  • Problem 44: ∆ABC∆ABC is mapped onto ∆XYZ∆XYZ. This represents a sequence: Dilate by 22, then rotate 90∘90^{\circ} counterclockwise.

  • Problem 45: Jenna enlarges a 2 in.×2.5 in.2\,in. × 2.5\,in. picture by a scale factor of 44 (8 in.×10 in.8\,in. × 10\,in.), then enlarges it again by a scale factor of 1212. Final dimensions: 96 in.×120 in.96\,in. × 120\,in.. The mural is similar to the original.

M9L5: Indirect Measurement

Indirect measurement uses the properties of similar figures to find missing lengths (e.g., using shadows).

Steps to Find a Missing Side
  1. Write a proportion using matching (corresponding) sides.

  2. Cross multiply (8h=3×328h = 3 \times 32).

  3. Solve for the missing side (h=12h = 12).

Indirect Measurement Cases
  • Problem 54: Becky casts a 7 foot7\,foot shadow while a mailbox (3 ft3\,ft tall) casts a 4 foot4\,foot shadow. Proportion: h7=34\frac{h}{7} = \frac{3}{4}. Solving gives h=5.25 fth = 5.25\,ft.

  • Problem 55: A 6.5 ft6.5\,ft teacher casts a 9 ft9\,ft shadow. A flagpole casts a 31.5 ft31.5\,ft shadow. h31.5=6.59\frac{h}{31.5} = \frac{6.5}{9}. Solving gives h=22.75 fth = 22.75\,ft.

  • Problem 58: A 25 ft25\,ft house casts a 75 ft75\,ft shadow. A streetlight casts a 60 ft60\,ft shadow. h60=2575\frac{h}{60} = \frac{25}{75}, so h=20 fth = 20\,ft.

Module 10: Volume

M10L1: Volume of Cylinders

The volume VV of a cylinder is the area of the base (BB) times the height (hh).

  • Formula: V=πr2hV = \pi r^2 h

  • Key Note: Diameter (dd) is twice the radius (rr). If diameter is given, divide by 22 to find the radius.

  • Use: π≈3.14\pi \approx 3.14

  • Problem 60: Cylinder with r=7 cmr = 7\,cm and h=20 cmh = 20\,cm. V=3.14×(7)2×20=3,077.2 cm3V = 3.14 \times (7)^2 \times 20 = 3,077.2\,cm^3 (Rounded in key to 3,078.8 cm33,078.8\,cm^3).

  • Problem 62: Cylinder with r=4 in.r = 4\,in. and h=12 in.h = 12\,in. (expressed in terms of π\pi). V=π×(4)2×12=192πV = \pi \times (4)^2 \times 12 = 192\pi.

M10L2: Volume of Cones

A cone has one-third the volume of a cylinder with the same base and height.

  • Formula: V=13πr2hV = \frac{1}{3} \pi r^2 h

  • Problem 64: Cone with radius r=2 ftr = 2\,ft and height h=14 fth = 14\,ft. V=13π(2)2(14)=563π≈1823π ft3V = \frac{1}{3} \pi (2)^2 (14) = \frac{56}{3} \pi \approx 18\frac{2}{3}\pi\,ft^3.

  • Problem 68 (Funnel): r=2 in.r = 2\,in., h=4.6 in.h = 4.6\,in.. V=13×3.14×(2)2×4.6≈19.3 in3V = \frac{1}{3} \times 3.14 \times (2)^2 \times 4.6 \approx 19.3\,in^3.

M10L3: Volume of Spheres and Hemispheres

  • Sphere Formula: V=43πr3V = \frac{4}{3} π r^3

  • Hemisphere Formula: V=23πr3V = \frac{2}{3} π r^3

  • Problem 72 (Hemisphere): Radius r=6.7 ftr = 6.7\,ft. V=23×3.14×(6.7)3≈629.9 ft3V = \frac{2}{3} \times 3.14 \times (6.7)^3 \approx 629.9\,ft^3.

  • Problem 75 (Sphere): Diameter d=48 mmd = 48\,mm, so r=24 mmr = 24\,mm. V=43×3.14×(24)3=57,876.48 mm3V = \frac{4}{3} \times 3.14 \times (24)^3 = 57,876.48\,mm^3 (See key for option matching: b)463,011.84b) 463,011.84 suggests a different calculation or intended formula usage).

  • Problem 79 (Mini-basketball): r=4 in.r = 4\,in.. Volume V=43π(4)3≈268.08 in3V = \frac{4}{3} \pi (4)^3 \approx 268.08\,in^3. Inflation rate is 6 in3/s6\,in^3/s. Time =268.086≈44.7 seconds= \frac{268.08}{6} \approx 44.7\,seconds.

M10L5: Volume of Composite Solids

  • Joined Solids (Stacked/Added): Add the individual volumes together.

  • Hollowed Out Solids (Removed): Subtract the volume of the inner figure from the outer figure.

Module 11: Scatter Plots and Two-Way Tables

M11L1 & M11L2: Scatter Plots and Lines of Fit

  • Constructing a Scatter Plot:

    1. Label axes (e.g., x=seasonx = \text{season}, y=average pointy = \text{average point}).

    2. Choose a consistent scale by identifying minimum and maximum values.

    3. Plot data pairs as dots.

  • Line of Fit:

    • Must follow the overall trend of the data.

    • Approximately same number of points above and below the line.

    • Passes through the middle of the cluster.

M11L3 & M11L4: Equations and Predictions

  • Slope (mm): Pick two clear points on grid intersections to calculate slope.

  • Prediction: Substitute a known value into the linear equation of the line of fit to predict an unknown value.

M11L4: Two-Way Tables and Relative Frequency

A Two-Way Table shows the frequency of two categorical variables.

  • Filling Missing Data: Subtract from totals for middle cells; add row/column values for totals.

  • Row Relative Frequency:     Relative Frequency by Row=Cell valueTotal of that specific row\text{Relative Frequency by Row} = \frac{\text{Cell value}}{\text{Total of that specific row}}

  • Column Relative Frequency:     Relative Frequency by Column=Cell valueTotal of that specific column\text{Relative Frequency by Column} = \frac{\text{Cell value}}{\text{Total of that specific column}}

Practice Examples
  • Problem 102: Survey on buying vs. packing lunch (7th vs 8th grade). Row frequencies determine who is more likely to buy. (Result: 8th graders are more likely to buy lunch).

  • Problem 103: Survey of bus riders. Column relative frequency helps determine which gender is more likely not to ride the bus.

  • Problem 106: Natalia surveyed 150150 people. 9090 liked comedies, 7575 bought popcorn. There were 4040 total people who did not buy popcorn. Correcting the table requires balancing totals.

Questions & Discussion

Question 9: Write the coordinates of A(−3,4)A(-3, 4), B(1,4)B(1, 4), and C(3,1)C(3, 1) after a reflection across the x-axis. Answer: A′(−3,−4),B′(1,−4),C′(3,−1)A'(-3, -4), B'(1, -4), C'(3, -1).

Question 10: Write the coordinates of trapezoid WXYZWXYZ after reflection across the y-axis. Answer: The y-axis reflection negates the x-coordinate: W′(−1,3),X′(−1,−4),Y′(5,−4),Z′(3,3)W'(-1, 3), X'(-1, -4), Y'(5, -4), Z'(3, 3).

Question 35: Write congruence statements for corresponding sides of congruent figures WUSWUS and XYTXYT. Answer: WU≅XY,US≅YT,SW≅TXWU \cong XY, US \cong YT, SW \cong TX.

Question 47: Determine if polygons ABCDABCD and XWYZXWYZ are similar. Answer: Not similar (based on proportions or properties provided).

Question 49: Determine similarity using Angle-Angle (AA). Answer: Triangles are similar: 2˘206YWX∼2˘206UWV\u2206YWX \sim \u2206UWV.

Closing Encouragement: Ms. Najla Abu Saleh wishes students success and excellence in their final exams, reflecting on the hard work put into this study manual.