D \(414 \text{ meters}\)
No Inside Lines: Perimeter only includes the two straight sides (\(2 \times 100\)) and the combined circumference of the two ends (\(\pi \times 50\)).
4-Function / Square Root: Set \(\pi \approx 3.14\). Calculate \(3.14 \times 50 = 157\). Add straight sides: \(157 + 200 = 357\).
TI-84 Precision: Use the literal 2*100 + π*50 to avoid pre-rounding errors.
Shape Shifter Math Slide 4 of 12
Q4
Surface Area & Volume
A cylinder has a radius of \(3 \text{ cm}\) and a height of \(10 \text{ cm}\). What is the total surface area of the cylinder in terms of \(\pi\)?
(Formula: \(SA = 2\pi r^2 + 2\pi r h\))
A \(60\pi \text{ cm}^2\)
B \(69\pi \text{ cm}^2\)
C \(78\pi \text{ cm}^2\)
D \(118\pi \text{ cm}^2\)
Keep \(\pi\) factored: Treat \(\pi\) like a variable. Sum the coefficients of both components.
Mental Math:
Base circles: \(2 \times (3^2) = 18\).
Side wrapper: \(2 \times 3 \times 10 = 60\).
Total: \(18 + 60 = 78\). Add \(\pi\) at the end.
4-Function Tip: Never press the \(\pi\) button if answers contain \(\pi\). Just calculate: 2*3*3 + 2*3*10.
Shape Shifter Math Slide 5 of 12
Q5
Surface Area & Volume
A storage cone has a radius of \(6 \text{ feet}\) and a slant height of \(10 \text{ feet}\). What is the volume of this storage cone?
(Formula: \(V = \frac{1}{3}\pi r^2 h\))
A \(96\pi \text{ ft}^3\)
B \(120\pi \text{ ft}^3\)
C \(288\pi \text{ ft}^3\)
D \(360\pi \text{ ft}^3\)
Find \(h\) first: Height \(h\) is not given. Use Pythagorean Theorem with slant height (\(10\)) and radius (\(6\)): \(h = \sqrt{10^2 - 6^2} = 8\).
Square Root Calc: Enter 100 - 36 to get \(64\). Press root key to get \(8\). Then calculate: \((1/3) \times 36 \times 8 = 96\).
TI-84 shortcut: Input (1/3) * (6^2) * √(10^2 - 6^2) to get \(96\) directly.
Shape Shifter Math Slide 6 of 12
Q6
Transformations
Point \(P(4, -3)\) is rotated \(90^\circ\) counterclockwise about the origin, and then reflected across the x-axis. What are the coordinates of the final point \(P''\)?
(Hint: Rotation \(90^\circ\) CCW is \((x, y) \rightarrow (-y, x)\))
A \((3, -4)\)
B \((3, 4)\)
C \((-3, -4)\)
D \((-4, 3)\)
Step 1 (Rotation): \((4, -3) \rightarrow (-(-3), 4) = (3, 4)\).
Step 2 (Reflection over X): Keep x same, negate y: \((x, y) \rightarrow (x, -y)\). So \((3, 4) \rightarrow (3, -4)\).
Mental Sketch: Imagine quadrant transformations. Original is IV. Rotate CCW 90 degrees → Quadrant I \((3, 4)\). Reflect over horizontal axis → Quadrant IV \((3, -4)\).
Shape Shifter Math Slide 7 of 12
Q7
Congruence & Similarity
A flagpole casts a shadow that is \(24 \text{ feet}\) long. At the same time, a yardstick (\(3\text{ feet}\) tall) standing vertically casts a shadow that is \(4 \text{ feet}\) long. What is the height of the flagpole?
A \(18 \text{ feet}\)
B \(20 \text{ feet}\)
C \(24 \text{ feet}\)
D \(32 \text{ feet}\)
Ratio Setup: \(\frac{\text{Flag Height}}{\text{Flag Shadow}} = \frac{\text{Yardstick Height}}{\text{Yardstick Shadow}}\)
\(\frac{h}{24} = \frac{3}{4}\).
Mental Scale Factor: Flagpole shadow (\(24\)) is \(6\) times longer than yardstick shadow (\(4\)). Scale up yardstick height: \(3 \times 6 = 18\).
4-Function cross-multiply: Key in 24 * 3 / 4.
Shape Shifter Math Slide 8 of 12
Q8
Right Triangle Trigonometry
A tourist stands \(100 \text{ feet}\) away from the base of a monument. Looking up, the tourist measures the angle of elevation to the top of the monument as \(35^\circ\). What is the height of the monument?
(Trig values: \(\sin(35^\circ) \approx 0.57\), \(\cos(35^\circ) \approx 0.82\), \(\tan(35^\circ) \approx 0.70\))
A \(57 \text{ feet}\)
B \(70 \text{ feet}\)
C \(82 \text{ feet}\)
D \(143 \text{ feet}\)
Choose the right ratio: Height is "opposite" side, distance is "adjacent". Opposite and Adjacent means we use Tangent!
Formula: \(\tan(35^\circ) = \frac{\text{Height}}{100}\)
\(\text{Height} = 100 \times \tan(35^\circ)\).
TI-84 Precision Check: Check mode is DEGREE first. Key in 100 * tan(35).
Shape Shifter Math Slide 9 of 12
Q9
Spatial Reasoning
A shipping box has dimensions of \(3 \text{ feet}\) by \(4 \text{ feet}\) by \(12 \text{ feet}\). What is the length of the longest rod that can fit completely inside this box from one corner to the opposite corner?
A \(12.5 \text{ feet}\)
B \(13 \text{ feet}\)
C \(15 \text{ feet}\)
D \(19 \text{ feet}\)
3D Pythagorean Theorem: Extend \(a^2 + b^2 = c^2\) into three dimensions: \(d = \sqrt{l^2 + w^2 + h^2}\).
Mental Math: Base has dimensions \(3\) and \(4\), forming a \(3-4-5\) right triangle. Its diagonal is \(5\). Now find hypotenuse of legs \(5\) and \(12\): a \(5-12-13\) triple!
Square Root Calc: Enter 3^2 + 4^2 + 12^2 = 169. Press root key → \(13\).
Shape Shifter Math Slide 10 of 12
Q10
Right Triangle Trigonometry
The height of an equilateral triangle is \(12 \text{ inches}\). What is the perimeter of the triangle in exact radical form?
A \(12\sqrt{3} \text{ inches}\)
B \(24 \text{ inches}\)
C \(24\sqrt{3} \text{ inches}\)
D \(36\sqrt{3} \text{ inches}\)
30-60-90 Decomposition: Splitting equilateral triangle in half creates two \(30^\circ-60^\circ-90^\circ\) triangles. Height is long leg \(x\sqrt{3}\).
Ratios: If long leg is \(12\), short leg \(x\) is \(\frac{12}{\sqrt{3}} = 4\sqrt{3}\). Hypotenuse (full side) is \(2x = 8\sqrt{3}\).
Perimeter: There are three equal sides, so: \(3 \times (8\sqrt{3}) = 24\sqrt{3}\).
Shape Shifter Math Slide 11 of 12
Session Checkpoint Ready for Practice
You are about to receive the Shape Shifter Practice Worksheet containing ten parallel practice problems.
Worksheet Rules: Show all your work, construct clear diagrams when helpful, and solve independently. No hints will be provided!
Geometric & Spatial Reasoning Section Complete
Time to Practice
Coordinate point \(A(-2, 5)\) is rotated \(90^\circ\) counterclockwise about the origin, and then the resulting point is reflected across the x-axis. What are the coordinates of the final point \(A''\)? (Transformation rule: \(90^\circ\) CCW rotation is \((x, y) \rightarrow (-y, x)\))
A) \((-5, 2)\) B) \((-5, -2)\) C) \((5, -2)\) D) \((5, 2)\)
Show Your Formulation Here:
TSIA2 Mathematics Module Workbook Page 3 of 5
07
A tall oak tree casts a shadow that is \(36 \text{ feet}\) long. At the exact same time of day, a high school student who is \(6 \text{ feet}\) tall casts a vertical shadow of \(9 \text{ feet}\). What is the height of the oak tree?
A) \(24 \text{ feet}\) B) \(27 \text{ feet}\) C) \(30 \text{ feet}\) D) \(54 \text{ feet}\)
Show Your Formulation Here:
08
A child's slide is installed at an angle of elevation of \(40^\circ\) with the ground. The distance from the base of the access ladder to the end of the slide's shadow along the ground is exactly \(8 \text{ feet}\). Find the vertical height of the slide. (Trigonometric values: \(\sin(40^\circ) \approx 0.64\); \(\cos(40^\circ) \approx 0.77\); \(\tan(40^\circ) \approx 0.84\))
A) \(5.12 \text{ feet}\) B) \(6.16 \text{ feet}\) C) \(6.72 \text{ feet}\) D) \(9.52 \text{ feet}\)
Show Your Formulation Here:
TSIA2 Mathematics Module Workbook Page 4 of 5
09
A rectangular shipping crate measures \(6 \text{ inches}\) wide, \(8 \text{ inches}\) deep, and \(24 \text{ inches}\) tall. What is the length of the longest solid steel rod that can fit entirely inside the crate from one base corner to the opposite top corner?
A) \(24.7 \text{ in}\) B) \(25.0 \text{ in}\) C) \(26.0 \text{ in}\) D) \(38.0 \text{ in}\)
Show Your Formulation Here:
10
The vertical altitude of an equilateral triangle is exactly \(18\sqrt{3} \text{ cm}\). What is the total perimeter of this equilateral triangle?
A) \(54 \text{ cm}\) B) \(72 \text{ cm}\) C) \(108 \text{ cm}\) D) \(108\sqrt{3} \text{ cm}\)
Show Your Formulation Here:
TSIA2 Mathematics Module Workbook Page 5 of 5
Q7: Proportions in Similar Triangles Answer: A (\(18 \text{ ft}\))
The flagpole and yardstick form similar right triangles with the ground. Setup the proportion: \[\frac{\text{Flagpole Height}}{\text{Flagpole Shadow}} = \frac{\text{Yardstick Height}}{\text{Yardstick Shadow}} \implies \frac{h}{24} = \frac{3}{4}\] Cross-multiply and solve: \(4h = 72 \implies h = 18 \text{ feet}\).
Q8: Trigonometric Applications Answer: B (\(70 \text{ ft}\))
Ground distance (\(100 \text{ ft}\)) is the Adjacent Leg. Height (\(h\) ) is the Opposite Leg. Use tangent: \[\tan(35^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} \implies \tan(35^\circ) = \frac{h}{100}\] Solve: \(h = 100 \times \tan(35^\circ) \approx 100 \times 0.70 = 70 \text{ feet}\).
Q9: The 3D Pythagorean Theorem Answer: B (\(13 \text{ ft}\))
The diagonal \(d\) of a rectangular box is found using: \[d = \sqrt{l^2 + w^2 + h^2}\] Substitute the given values \(l=3, w=4, h=12\): \[d = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13 \text{ feet}\].
Q10: Special Right Triangles Answer: C (\(24\sqrt{3} \text{ in}\))
Altitude of an equilateral triangle forms two \(30^\circ-60^\circ-90^\circ\) triangles. The altitude is \(x\sqrt{3}\): \[x\sqrt{3} = 12 \implies x = \frac{12}{\sqrt{3}} = 4\sqrt{3} \text{ in}\] Equilateral side is hypotenuse (\(2x\)): \(\text{Side} = 8\sqrt{3} \text{ in}\). Perimeter: \(P = 3 \times \text{Side} = 3 \times 8\sqrt{3} = 24\sqrt{3} \text{ inches}\).
TSIA2 Geometric & Spatial Reasoning Key Page 2 of 4
TSIA2 Solutions Reference
Q1C
Q2B
Q3C
Q4C
Q5A
Q6A
Q7A
Q8C
Q9C
Q10C
Q1: Speed Conversion (90 mph to ft/s) Answer: C (\(132 \text{ ft/s}\))
Employ standard dimensional analysis: \[90 \text{ miles/hour} \times \frac{5,280 \text{ feet}}{1 \text{ mile}} \times \frac{1 \text{ hour}}{3,600 \text{ seconds}}\] Multiply: \(\frac{90 \times 5280}{3600} = \frac{475,200}{3600} = 132 \text{ feet per second}\).
Q2: Area Conversion (540 sq ft to sq yd) Answer: B (\(60 \text{ sq yd}\))
Using squared conversion (\(1 \text{ sq yd} = 3 \text{ ft} \times 3 \text{ ft} = 9 \text{ sq ft}\)): \[540 \text{ sq ft} \times \frac{1 \text{ sq yd}}{9 \text{ sq ft}} = \frac{540}{9} = 60 \text{ square yards}\].
Q3: Composite Perimeter (Garden with Semicircles) Answer: C (\(143 \text{ ft}\))
Perimeter features the two horizontal rectangle sides (\(2 \times 40 = 80 \text{ ft}\)) and the combined semicircular boundaries (diameter \(d = 20 \text{ ft}\)): \[P = 80 + \pi \times 20 \approx 80 + (3.14159 \times 20) = 80 + 62.83 = 142.83 \approx 143 \text{ feet}\].
Q4: Cylinder Surface Area (r=5, h=12) Answer: C (\(170\pi \text{ sq in}\))
Substitute \(r=5\) and \(h=12\) into surface area formula: \[SA = 2\pi r^2 + 2\pi r h \implies SA = 2\pi (5^2) + 2\pi (5)(12) = 50\pi + 120\pi = 170\pi \text{ sq in}\].
Q5: Cone Volume Capacity (r=9, slant=15) Answer: A (\(324\pi \text{ ft}^3\))
Determine vertical height \(h\) via Pythagorean Theorem: \(h = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12 \text{ ft}\). Now insert into volume formula: \[V = \frac{1}{3}\pi r^2 h \implies V = \frac{1}{3}\pi (9^2)(12) = \frac{1}{3}\pi (81)(12) = 324\pi \text{ ft}^3\].
TSIA2 Geometric & Spatial Reasoning Key Page 3 of 4
TSIA2 Solutions Reference
Q6: Sequential Coordinate Transformations Answer: A (\((-5, 2)\))
Apply the coordinates shifts sequentially for point \(A(-2, 5)\):
Step 1: Rotate \(90^\circ\) CCW: \((x, y) \rightarrow (-y, x)\). This gives \(A' = (-5, -2)\).
Step 2: Reflect across the x-axis: \((x, y) \rightarrow (x, -y)\). This yields \(A'' = (-5, 2)\).
Q7: Proportional Shadows Answer: A (\(24 \text{ ft}\))
Establish a similar-triangles shadow proportion: \[\frac{\text{Tree Height}}{\text{Tree Shadow}} = \frac{\text{Student Height}}{\text{Student Shadow}} \implies \frac{h}{36} = \frac{6}{9}\] Simplify \(\frac{6}{9}\) to \(\frac{2}{3}\). Solve for \(h\): \(h = 36 \times \frac{2}{3} = 24 \text{ feet}\).
Q8: Right Triangle Trigonometric Ratios Answer: C (\(6.72 \text{ ft}\))
The ladder base to shadow tip forms the horizontal leg (\(\text{Adjacent} = 8 \text{ ft}\)). The vertical height of the slide is the \(\text{Opposite}\) leg. The tangent ratio matches opposite and adjacent: \[\tan(40^\circ) = \frac{\text{Opposite}}{\text{Adjacent}} \implies \tan(40^\circ) = \frac{h}{8}\] Isolate the height: \(h = 8 \times \tan(40^\circ) \approx 8 \times 0.84 = 6.72 \text{ feet}\).
Q9: 3D Pythagorean Theorem Answer: C (\(26.0 \text{ in}\))
Use the 3D space diagonal equation with dimensions \(l=6, w=8, h=24\): \[d = \sqrt{l^2 + w^2 + h^2} = \sqrt{6^2 + 8^2 + 24^2} = \sqrt{36 + 64 + 576} = \sqrt{676} = 26 \text{ inches}\].
Q10: Special Right Triangles Answer: C (\(108 \text{ cm}\))
The altitude of an equilateral triangle partitions it into two \(30^\circ-60^\circ-90^\circ\) triangles. The altitude matches the long leg: \[x\sqrt{3} = 18\sqrt{3} \implies x = 18 \text{ cm} \text{ (the short leg)}\] The full side of the equilateral triangle is the hypotenuse (\(2x = 36 \text{ cm}\)). Compute perimeter: \(P = 3 \times \text{Side} = 3 \times 36 = 108 \text{ cm}\).
TSIA2 Geometric & Spatial Reasoning Key Page 4 of 4
