Year11 MATH 2-2-2 Warm-up Questions-Shape and Measurement (Spatial Intelligence).

To master Chapter 2 of General Mathematics, you must move beyond basic formulas and understand the proportional relationships between dimensions, area, and volume. This warm-up focuses on spatial reasoning, similarity, and the geometry of complex solids.

Concepts and Skills Covered:

1. Spherical Geometry: Applying volume ($V = \frac{4}{3}\pi r^3$) and surface area ($SA = 4\pi r^2$) formulas.
2. Cones and Pyramids: Using $V = \frac{1}{3}Ah$ and understanding the role of slant height in surface area.
3. Similarity and Scaling: Understanding that if lengths scale by $k$, areas scale by $k^2$ and volumes by $k^3$.
4. Composite Solids: Deconstructing complex shapes into standard geometric components.
5. Formula Manipulation: Solving for dimensions (like radius or height) when the total volume or area is known.

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Warm-up: Shape and Measurement (Spatial Intelligence)

Q1. A sphere has a radius of $3\text{ cm}$. Calculate its volume in terms of $\pi$.

A. 108π cm3

B. 27π cm3

C. 12π cm3

D. 36π cm3

Hint: The formula for the volume of a sphere is $V = \frac{4}{3}\pi r^3$.

Q2. A right cone has a radius of $5\text{ cm}$ and a slant height of $13\text{ cm}$. What is its total surface area?

A. 65π cm2

B. 115π cm2

C. 155π cm2

D. 90π cm2

Hint: Add the area of the circular base to the curved surface area, πr2+πrl.

Q3. A square-based pyramid has a base side length of $6\text{ m}$ and a vertical height of $10\text{ m}$. Calculate its volume.

A. 180 m3

B. 120 m3

C. 60 m3

D. 360 m3

Hint: The volume of any pyramid is one-third the volume of a prism with the same base and height.

Q4. Two similar storage bins have a linear scale factor of $k = 4$. If the smaller bin holds $2\text{ L}$ of grain, how much does the larger bin hold?

A. 8 L

B. 128 L

C. 16 L

D. 32 L

Hint: Remember that if lengths are multiplied by k, the volume is multiplied by k3.

Q5. The surface area of a sphere is $100\pi\text{ cm}^2$. What is its radius?

A. 10 cm

B. 2.5 cm

C. 5 cm

D. 25 cm

Hint: Equate the given area to the formula 4πr2 and solve for r.

Q6. Two similar triangles have areas of $10\text{ cm}^2$ and $90\text{ cm}^2$. What is the linear scale factor $k$ between them?

A. 3

B. 81

C. 4.5

D. 9

Hint: The ratio of the areas is equal to the square of the linear scale factor (k2).

Q7. A cylinder and a cone have the same radius and the same vertical height. If the volume of the cone is $50\text{ cm}^3$, what is the volume of the cylinder?

A. 150 cm3

B. 200 cm3

C. 100 cm3

D. 50 cm3

Hint: Compare the formulas Vcyl​=πr2h and Vcone​=31​πr2h.

Q8. Calculate the total surface area of a closed hemisphere with a radius of $10\text{ cm}$.

A. 300π cm2

B. 400π cm2

C. 200π cm2

D. 150π cm2

Hint: Don’t forget to include the area of the flat circular base (SAtotal​=3πr2).

Q9. If you double the radius of a cylinder while keeping its height the same, by what factor does the volume increase?

A. 16

B. 4

C. 2

D. 8

Hint: Look at the exponent of the radius in the cylinder volume formula.

Q10. A composite solid is made by placing a cone of height $4\text{ cm}$ on top of a cylinder of height $10\text{ cm}$. Both have a radius of $3\text{ cm}$. What is the total volume?

A. 102π cm3

B. 94π cm3

C. 126π cm3

D. 114π cm3

Hint: Calculate the volume of the cylinder and the cone separately, then add them together.

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Great work! Understanding how dimensions scale and interact in composite solids is the foundation for the more complex 3D modeling you’ll do in your PSMT. Keep up the momentum!

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一般数学の第2章をマスターするには、基本的な公式を超えて、寸法、面積、体積の比例関係を理解する必要があります。このウォームアップでは、空間的推論、相似、複雑な立体の幾何学に焦点を当てます。

カバーされる概念とスキル：

球面幾何学：体積（$V = \frac{4}{3}\pi r^3$）と表面積（$SA = 4\pi r^2$）の公式を適用します。

円錐とピラミッド：$V = \frac{1}{3}Ah$を使用し、表面積における斜高の役割を理解します。

相似とスケーリング：長さが$k$、面積が$k^2$、体積が$k^3$でスケーリングされることを理解します。

複合立体：複雑な形状を標準的な幾何学的構成要素に分解します。

数式操作：総体積または総面積が既知の場合、寸法（半径や高さなど）を解きます。

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Right answers.

Q1. D $36\pi\text{ cm}^3$

Using $V = \frac{4}{3}\pi r^3$, we get $\frac{4}{3} \times \pi \times 3^3 = \frac{4}{3} \times 27\pi = 36\pi$.

Q2. D $90\pi\text{ cm}^2$

Total surface area is the sum of the base ($\pi r^2 = 25\pi$) and the curved surface ($\pi rl = 65\pi$).

Q3. B $120\text{ m}^3$

The volume is $\frac{1}{3} \times \text{base area} \times \text{height} = \frac{1}{3} \times 36 \times 10 = 120$.

Q4. B $128\text{ L}$

Volume scales by $k^3$. $4^3 = 64$, so the new volume is $2 \times 64 = 128$.

Q5. C $5\text{ cm}$

Setting $4\pi r^2 = 100\pi$ leads to $r^2 = 25$, so $r = 5$.

Q6. A. $3$

The area scale factor is $90 / 10 = 9$. Since area scales by $k^2$, $k = \sqrt{9} = 3$.

Q7. A. $150\text{ cm}^3$

A cylinder’s volume is exactly three times that of a cone with the same dimensions.

Q8. A. $300\pi\text{ cm}^2$

The surface area is the sum of the curved hemisphere ($2\pi r^2 = 200\pi$) and the flat circular base ($\pi r^2 = 100\pi$).

Q9. B. $4$

The volume formula $\pi r^2 h$ involves $r^2$. Doubling $r$ means $(2r)^2 = 4r^2$.

Q10. A. $102\pi\text{ cm}^3$

Cylinder volume is $9\pi \times 10 = 90\pi$. Cone volume is $\frac{1}{3} \times 9\pi \times 4 = 12\pi$. Total is $102\pi$.

素晴らしい成果です！複合ソリッドにおける寸法のスケールと相互作用を理解することは、PSMTで行うより複雑な3Dモデリングの基礎となります。この勢いを維持してください！