Measures, Shape & Space
Mensuration in 3D
Find volumes and surface areas of prisms, cylinders, pyramids, cones, and spheres.
Mensuration in 3D focuses on volume (space inside solids) and surface area (total outer area).
Units and interpretation
- Length: cm, m
- Area: cm\(^2\), m\(^2\)
- Volume: cm\(^3\), m\(^3\)
Always keep units consistent before substituting values.
Prisms and cylinders
For any prism/cylinder:
\[ \text{Volume}=\text{base area}\times\text{height}. \]
Cylinder formulas:
\[ V=\pi r^2h,\qquad \text{TSA}=2\pi r^2+2\pi rh. \]
Pyramids and cones
\[ V_{\text{pyramid}}=\frac13(\text{base area})h,\qquad V_{\text{cone}}=\frac13\pi r^2h. \]
Cone curved surface area:
\[ \text{CSA}_{\text{cone}}=\pi rl \]
where \(l\) is slant height. Total surface area adds base area \(\pi r^2\).
Spheres
\[ V=\frac43\pi r^3,\qquad \text{Surface area}=4\pi r^2. \]
Composite solids
Split complex shapes into known solids, then add/subtract volumes or areas appropriately.
- Add volumes when parts are joined.
- Subtract volumes for holes/cut-outs.
- For surface area, remove hidden contact faces.
Volume in context
Common contexts include capacity, material usage, and mass:
\[ \text{mass}=\text{density}\times\text{volume}. \]
Exam strategy
- Sketch and label all dimensions before formula use.
- Choose exact \(\pi\) form first; round at final step if required.
- Convert units early (e.g., cm to m).
- For composite shapes, explain add/subtract logic briefly.
History
Mensuration developed from practical needs in architecture, agriculture, and trade. Ancient mathematicians studied area and volume formulas, with later refinement by Archimedes and others.
Derivation and reasoning
Why pyramid and cone have one-third factor
A pyramid (or cone) with same base area and height as a prism (or cylinder) has one-third the volume. This can be shown by dissection arguments or calculus in advanced courses.
Checkpoints
Find volume of a cylinder with \(r=3\) cm, \(h=10\) cm.
Answer: \(V=\pi r^2h=90\pi\) cm\(^3\).
Find total surface area of a closed cylinder with \(r=2\) cm, \(h=7\) cm.
Answer: \(2\pi r^2+2\pi rh=8\pi+28\pi=36\pi\) cm\(^2\).
Find volume of a cone with \(r=6\) cm, \(h=9\) cm.
Answer: \(\frac13\pi r^2h=\frac13\pi(36)(9)=108\pi\) cm\(^3\).
Sphere has radius 5 cm. Find its volume.
Answer: \(V=\frac43\pi(5^3)=\frac{500}{3}\pi\) cm\(^3\).
A block is \(4\times3\times2\) m. Density is \(2500\) kg/m\(^3\). Find mass.
Answer: Volume \(=24\) m\(^3\). Mass \(=2500\times24=60000\) kg.
Applications
- Construction: concrete volume and paint area estimation.
- Packaging: minimizing material for fixed volume containers.
- Engineering: tank capacity and material mass calculations.
References
- Standard high-school mensuration units on 3D solids.
- Geometry texts on prism, cylinder, cone, pyramid, and sphere formulas.
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