3D Solids: Volumes and Surface Areas
- Right Circular Cone:
where slant height $l = \sqrt{r^2 + h^2}$.Volume: $$V = \frac{1}{3}\pi r^2 h$$Total surface area: $$A = \pi r l + \pi r^2$$Curved surface area: $$\pi r l$$
- Sphere: $$V = \frac{4}{3}\pi r^3, \quad A = 4\pi r^2$$
- Right Pyramid: $$V = \frac{1}{3} \times \text{Base Area} \times h$$
- Angle Between a Line and a Plane: The angle between a line segment and its orthogonal projection onto the plane, computed via right-angled trigonometry $\tan \theta = \frac{\text{vertical height}}{\text{projected base length}}$.
Coordinates, Distances and Midpoints
- For $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$, $AB=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. Apply Pythagoras first in a horizontal plane, then vertically.
- The midpoint is $M(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2},\frac{z_1+z_2}{2})$. Average corresponding coordinates.
- Worked example: $A(1,2,0)$, $B(4,6,12)$ give $AB=\sqrt{3^2+4^2+12^2}=13$ and $M(2.5,4,6)$.
Space Diagonals and Line–Plane Angles
- A cuboid with perpendicular dimensions $a,b,h$ has base diagonal $p=\sqrt{a^2+b^2}$ and space diagonal $D=\sqrt{a^2+b^2+h^2}$.
- The angle between the space diagonal and the base is found in the right triangle with sides $p,h,D$: $\tan\theta=h/p$ or $\sin\theta=h/D$. A face diagonal is usually not the projection onto the base.
- Worked example: Dimensions $3,4,12$ give $p=5$, $D=13$ and $\theta=\tan^{-1}(12/5)\approx67.4^\circ$.
- When the line is parallel to the plane its angle is $0^\circ$; when perpendicular it is $90^\circ$.
Choosing a Solid Formula
| Solid | Volume | Surface area |
|---|---|---|
| Cuboid | $abh$ | $2(ab+ah+bh)$ |
| Right prism | Cross-section area × length | Two ends + exposed lateral faces |
| Closed cylinder | $\pi r^2h$ | $2\pi rh+2\pi r^2$ |
| Hemisphere | $\frac23\pi r^3$ | Curved $2\pi r^2$; with base $3\pi r^2$ |
| Right cone | $\frac13\pi r^2h$ | Curved $\pi rl$; with base $\pi rl+\pi r^2$ |
- Volume uses perpendicular height. Cone surface area uses slant height $l$; do not interchange them.
- For a pyramid, add the areas of the triangular faces separately. Face slant height differs from a sloping edge.
- Worked example: A square pyramid has base side $6$ and vertical height $4$. Its face slant height is $\sqrt{4^2+3^2}=5$. Thus $V=\frac13(36)(4)=48$ and total surface area $=36+4(\frac12\cdot6\cdot5)=96$.
Composite Solids, Similarity and Units
- Add volumes of non-overlapping parts; subtract cavities. Shared internal faces do not contribute to external surface area.
- With linear scale factor $k>0$, areas multiply by $k^2$ and volumes by $k^3$.
- Worked example: A cone filled to half its perpendicular height contains $1/8$ of its full volume: similar cones have height ratio $1/2$, radius ratio $1/2$, and volume ratio $(1/2)^3$.
- Use one unit system before substituting: $1\text{ m}=100\text{ cm}$, $1\text{ m}^2=10^4\text{ cm}^2$, $1\text{ m}^3=10^6\text{ cm}^3$; $1$ litre $=1000\text{ cm}^3$.
- Keep full calculator precision until the final answer. Check whether dimensions are radii or diameters, and whether a container is open or closed.
10 Progressive Practice Exercises
Work from single-step calculations to spatial and composite-solid applications.
Answer: $188\text{ cm}^2$
$V=6(8)(10)=480\text{ cm}^3$. $A=2(48+60+80)=376\text{ cm}^2$. $D=\sqrt{36+64+100}=\sqrt{200}\approx14.1\text{ cm}$.
Coordinate differences are $(4,3,12)$, so $AB=\sqrt{16+9+144}=13$. The midpoint is $(3,-0.5,9)$.
$V=\pi(3)^2(8)=72\pi\text{ cm}^3$. Both circular ends count: $A=2\pi(3)(8)+2\pi(3)^2=66\pi\text{ cm}^2$.
$\frac43\pi r^3=36\pi$ gives $r^3=27$, hence $r=3$ cm. Its area is $4\pi(3)^2=36\pi\text{ cm}^2$.
Base area $=36$. The face slant height is $\sqrt{4^2+3^2}=5$. $V=48\text{ cm}^3$, and $A=36+4(\frac12\cdot6\cdot5)=96\text{ cm}^2$.
The base projection has length $\sqrt{6^2+8^2}=10$ m. Thus $\theta=\tan^{-1}(12/10)\approx50.2^\circ$. Using $6$ or $8$ as the horizontal length would give a face angle instead.
Similarity gives water radius $5(6/12)=2.5$ cm. $V=\frac13\pi(2.5)^2(6)=12.5\pi\text{ cm}^3\approx39.3\text{ cm}^3$. This is $1/8$ of the full volume $100\pi$.
$V=90\pi+\frac23\pi(3)^3=108\pi\text{ cm}^3$. The joined circular face is internal. Exposed area $=2\pi(3)(10)+2\pi(3)^2+\pi(3)^2=87\pi\text{ cm}^2$.
Sphere volume $=36\pi\text{ cm}^3$. Each cone has volume $\frac13\pi(1)^2(3)=\pi\text{ cm}^3$. Hence $36$ complete cones. This assumes no material is lost and density is unchanged.
3D Geometry Formula Reference
Checks Before Submitting
- Choose the perpendicular height for volume and slant height for a cone’s curved surface.
- Count only exposed faces; label squared or cubed units.
- For an angle to a plane, identify the line’s projection onto that plane.
- Use degree mode for these angle calculations; preserve precision until the final result.
Exam Mini Test (5 Questions)
1. A sphere has a radius of $3.0\text{ cm}$. What is its volume?
2. A cuboid has perpendicular dimensions 3, 4 and 12. What is its space diagonal?
3. Convert 12 500 cubic centimetres to litres.
4. A sphere’s radius doubles. What happens to its volume?
5. A diagonal rises 12 m above a base projection of 10 m. Which expression gives its angle to the base?