3D shapes — cube, cuboid, cylinder, cone, sphere properties and formulas

Question

A solid cylinder of radius 7 cm and height 10 cm is melted and recast into cones of radius 3.5 cm and height 4 cm. How many cones can be made?

Solution — Step by Step

V_{cylinder} = \pi r^2 h = \pi \times 7^2 \times 10 = 490\pi \text{ cm}^3

V_{cone} = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \times 3.5^2 \times 4 = \frac{1}{3}\pi \times 12.25 \times 4 = \frac{49\pi}{3} \text{ cm}^3

Since metal is conserved (melted and recast), total volume stays the same.

n = \frac{V_{cylinder}}{V_{cone}} = \frac{490\pi}{49\pi/3} = \frac{490 \times 3}{49} = \frac{1470}{49} = \mathbf{30 \text{ cones}}

Why This Works

graph TD
    A["3D Shape Formula Selection"] --> B["Cube: all sides equal"]
    A --> C["Cuboid: length ≠ breadth ≠ height"]
    A --> D["Cylinder: circular cross-section"]
    A --> E["Cone: circular base, pointed top"]
    A --> F["Sphere: all points equidistant from centre"]
    B --> G["V = a³, TSA = 6a²"]
    C --> H["V = lbh, TSA = 2 lb+bh+hl"]
    D --> I["V = πr²h, TSA = 2πr r+h"]
    E --> J["V = ⅓πr²h, TSA = πr r+l"]
    F --> K["V = ⅘πr³, TSA = 4πr²"]

The key principle in “melting and recasting” problems: volume is conserved. The shape changes but the amount of material stays the same. So we equate the volume of the original solid with the total volume of all the new solids.

Shape	Volume	Lateral/Curved SA	Total SA
Cube ( $a$ )	$a^3$	$4a^2$	$6a^2$
Cuboid ( $l,b,h$ )	$lbh$	$2h(l+b)$	$2(lb+bh+hl)$
Cylinder ( $r,h$ )	$\pi r^2 h$	$2\pi rh$	$2\pi r(r+h)$
Cone ( $r,h,l$ )	$\frac{1}{3}\pi r^2 h$	$\pi rl$	$\pi r(r+l)$
Sphere ( $r$ )	$\frac{4}{3}\pi r^3$	$4\pi r^2$	$4\pi r^2$
Hemisphere ( $r$ )	$\frac{2}{3}\pi r^3$	$2\pi r^2$	$3\pi r^2$

Note: for a cone, $l = \sqrt{r^2 + h^2}$ (slant height).

Alternative Method

For “how many small shapes from one big shape” problems, the calculation is always: $n = V_{big} / V_{small}$ . Keep $\pi$ in both expressions — it cancels every time. Many students compute decimal values of both volumes separately and then divide, introducing rounding errors. Keeping $\pi$ symbolic is cleaner and faster.

Common Mistake

Confusing surface area conservation with volume conservation. When a solid is melted and recast, only VOLUME is conserved — surface area changes. A sphere melted into a thin wire has the same volume but vastly more surface area. Students sometimes set up equations equating surface areas instead of volumes in recasting problems. Always ask: is the material being reshaped (volume conserved) or being painted/wrapped (surface area relevant)?

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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