• Actual Area: \(\text{Area} = 80\text{ m} \times 120\text{ m} = \mathbf{9,600\text{ m}^2}\)
Common Misconception: Students often calculate map area \((4 \times 6 = 24\text{ cm}^2)\) and multiply by 20 \((480\text{ m}^2)\). Emphasize multiplying by \(20^2 = 400\) \((24 \times 400 = 9,600\text{ m}^2)\).
Question 8 Solution: Blueprint Room (8.4 cm × 5 cm, Scale: 1 cm = 2.5 m) 21 m × 12.5 m | 262.5 m²
• Dimensions: \(\text{Length} = 8.4\text{ cm} \times 2.5\text{ m/cm} = \mathbf{21\text{ meters}}\); \(\text{Width} = 5\text{ cm} \times 2.5\text{ m/cm} = \mathbf{12.5\text{ meters}}\)
• Actual Area: \(\text{Area} = 21\text{ m} \times 12.5\text{ m} = \mathbf{262.5\text{ m}^2}\)
Question 9 Solution: Rectangular Park (2.5 in × 4 in, Scale: 1 in = 40 m) 100 m × 160 m | 16,000 m²
• Dimensions: \(\text{Width} = 2.5\text{ in} \times 40\text{ m/in} = \mathbf{100\text{ meters}}\); \(\text{Length} = 4\text{ in} \times 40\text{ m/in} = \mathbf{160\text{ meters}}\)
• Actual Area: \(\text{Area} = 100\text{ m} \times 160\text{ m} = \mathbf{16,000\text{ m}^2}\)
Question 10 Solution: Scale Area Relationship (Linear scale 1 in = 8 ft, Area = 12 in²) Correct: C (768 sq ft)
• Linear scale factor: \(k = 8\text{ ft/in}\)
• Area scale factor: \(k^2 = 8^2 = \mathbf{64\text{ sq ft / sq in}}\)
• Actual Area: \(12\text{ in}^2 \times 64\frac{\text{sq ft}}{\text{in}^2} = \mathbf{768\text{ sq ft}}\)
Distractor Analysis: A (96) multiplies by 8 once (linear trap); B (384) multiplies by 32; D (1,536) doubles the correct area.
Question 11 Solution: Two-Map Conversion (Map 1: 1 in = 20 mi; Map 2: 1 in = 15 mi) Correct: C (6.0 inches)
• Step 1 (Actual distance): \(4.5\text{ in} \times 20\text{ mi/in} = \mathbf{90\text{ miles}}\)
• Step 2 (Map 2 distance): \(\frac{90\text{ miles}}{15\text{ mi/in}} = \mathbf{6.0\text{ inches}}\)
Distractor Analysis: A (3.375) divides by 20 then multiplies by 15; B (5.25) arithmetical error; D (6.75) sets up inverse ratio incorrectly.
TSIA2 Instructor Solution Guide • Page 2: 2D Scale Area & Standardized Assessment (Q7–Q11)
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Section 3 Key 2D Plane Geometry & 3D Volume Foundations (Q12–Q21)
Items: Q12–Q21
2D Plane Geometry Solutions (Q12–Q15) Triangles, Circles & Rectangles
Q12: Triangle Area (\(b=12, h=9\)) 54 cm²
\(A = \frac{1}{2}bh = \frac{1}{2}(12)(9) = 6 \times 9 = \mathbf{54\text{ cm}^2}\)
Q13: Circle Area (\(r=5\text{ ft}\)) 25π ft²
\(A = \pi r^2 = \pi (5)^2 = \mathbf{25\pi\text{ ft}^2}\) \((\approx 78.54\text{ ft}^2)\)
Q14: Rectangle Length (\(A=72, w=6\)) 12 meters
\(A = lw \implies l = \frac{A}{w} = \frac{72}{6} = \mathbf{12\text{ meters}}\)
Q15: Circle Radius (\(A=49\pi\)) Correct: A (7)
\(\pi r^2 = 49\pi \implies r^2 = 49 \implies r = \sqrt{49} = \mathbf{7}\)
3D Volume & Measurement Solutions (Q16–Q21) Prisms, Cubes & Cylinders
Q16: Prism Volume (\(8 \times 3 \times 5\)) 120 in³
\(V = lwh = 8 \times 3 \times 5 = 24 \times 5 = \mathbf{120\text{ in}^3}\)
Q17: Cube Side Length (\(V = 125\text{ cm}^3\)) 5 cm
\(V = s^3 \implies s = \sqrt[3]{125} = \mathbf{5\text{ cm}}\)
Q18: Cylinder Volume (\(r=4, h=10\)) 160π cm³
\(V = \pi r^2 h = \pi (4)^2 (10) = 16\pi \times 10 = \mathbf{160\pi\text{ cm}^3}\)
Q19: Prism Volume (\(B=24\text{ ft}^2, h=7\)) 168 ft³
\(V = Bh = 24\text{ ft}^2 \times 7\text{ ft} = \mathbf{168\text{ ft}^3}\)
Q20: Cylinder MC (\(r=3, h=8\)) Correct: B (72π)
\(V = \pi (3)^2 (8) = 9\pi \times 8 = \mathbf{72\pi}\). Distractor A forgets to square \(r\) \((24\pi)\).
Q21: Prism Height MC (\(V=120, B=30\)) Correct: C (4 cm)
\(V = Bh \implies h = \frac{120}{30} = \mathbf{4\text{ cm}}\)
TSIA2 Instructor Solution Guide • Page 3: 2D Geometry & 3D Volume Foundations (Q12–Q21)
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Section 4 Key Complex Solids, Error Analysis & Design Optimization (Q22–Q27)
Items: Q22–Q27
Q22: Triangular Prism 360 cm³
• \(B = \frac{1}{2}(10)(6) = 30\text{ cm}^2\)
• \(V = B \times L = 30 \times 12 = \mathbf{360\text{ cm}^3}\)
Q23: Cone Volume 27π cm³
• \(V = \frac{1}{3}\pi (3)^2 (9)\)
• \(V = \frac{1}{3}\pi (9)(9) = \mathbf{27\pi\text{ cm}^3}\)
Q24: Sphere Volume \(\frac{32}{3}\pi\) cm³
• \(V = \frac{4}{3}\pi (2)^3 = \frac{4}{3}\pi (8)\)
• \(V = \mathbf{\frac{32}{3}\pi\text{ cm}^3} \approx 10.67\pi\)
Question 25 Solution: Conceptual Error Proof (Doubling Cylinder Radius) Claim is INCORRECT (No)
• Original Volume: \(V_1 = \pi r^2 h\)
• New Volume with \(r_{new} = 2r\): \(V_2 = \pi (2r)^2 h = \pi (4r^2) h = 4(\pi r^2 h) = \mathbf{4 V_1}\)
Instructional Note / Rubric: Doubling the radius quadruples (\(\times 4\)) the volume, because radius is raised to the second power. Full credit requires: (1) stating “No/Incorrect”, (2) showing \((2r)^2 = 4r^2\), and (3) concluding volume multiplies by 4, not 2.
Question 26 Solution: Surface Area Minimization (\(V = 240\text{ cm}^3\)) Min SA = 236 cm² (\(5 \times 6 \times 8\))
• Set A Example: \(4\text{ cm} \times 6\text{ cm} \times 10\text{ cm}\) → \(SA = 2(24 + 60 + 40) = 2(124) = \mathbf{248\text{ cm}^2}\)
• Set B Example (Optimal): \(5\text{ cm} \times 6\text{ cm} \times 8\text{ cm}\) → \(SA = 2(30 + 48 + 40) = 2(118) = \mathbf{236\text{ cm}^2}\)
Optimization Principle: For a fixed volume, surface area is minimized when the prism is as close to a cube as possible (\(\sqrt[3]{240} \approx 6.21\text{ cm}\)). The integer dimensions closest to 6.21 are \(5 \times 6 \times 8\), which yields the absolute lowest integer surface area (\(236\text{ cm}^2\)).
Question 27 Solution: Packaging Engineering Challenge (\(V \ge 500\text{ in}^3\)) Rubric & Exemplar
• Model Exemplar 1 (Near-Cube): \(8\text{ in} \times 8\text{ in} \times 8\text{ in} = \mathbf{512\text{ in}^3}\); \(\text{Surface Area} = 6(8^2) = \mathbf{384\text{ sq in}}\)
• Model Exemplar 2 (Standard Box): \(10\text{ in} \times 10\text{ in} \times 5\text{ in} = \mathbf{500\text{ in}^3}\); \(\text{Surface Area} = 2(100 + 50 + 50) = \mathbf{400\text{ sq in}}\)
Efficiency Criteria: Full credit requires: (1) \(l \times w \times h \ge 500\text{ in}^3\), (2) valid surface area calculation, and (3) justification explaining that a near-cubic form factor minimizes sheet material / manufacturing cost while maximizing structural integrity and stacking efficiency.
TSIA2 Instructor Solution Guide • Page 4: Complex Solids, Proofs & Optimization (Q22–Q27)
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