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

Question

A solid cylinder has radius $r = 7$ cm and height $h = 10$ cm. Find its total surface area and volume. If this cylinder is melted and recast into a cone of the same radius, find the height of the cone.

(CBSE 9 & 10 — surface areas and volumes)

Solution — Step by Step

Total Surface Area = $2\pi r(r + h)$ (two circular faces + curved surface)

TSA = 2 \times \frac{22}{7} \times 7 \times (7 + 10) = 2 \times 22 \times 17 = \mathbf{748 \text{ cm}^2}

Volume = $\pi r^2 h$

V = \frac{22}{7} \times 49 \times 10 = 22 \times 70 = \mathbf{1540 \text{ cm}^3}

When melted and recast, the volume stays the same (no material is lost or gained).

V_{\text{cylinder}} = V_{\text{cone}}

\pi r^2 h_{\text{cyl}} = \frac{1}{3}\pi r^2 h_{\text{cone}}

Since $r$ is the same, cancel $\pi r^2$ :

h_{\text{cyl}} = \frac{1}{3}h_{\text{cone}}

h_{\text{cone}} = 3 \times h_{\text{cyl}} = 3 \times 10 = \mathbf{30 \text{ cm}}

Why This Works

The cone has $1/3$ the volume of a cylinder with the same base and height. So to hold the same material, the cone must be 3 times as tall.

graph TD
    A["3D Shape Problem"] --> B{"What's asked?"}
    B -->|"Surface area"| C{"Which shape?"}
    B -->|"Volume"| D{"Which shape?"}
    B -->|"Melting/recasting"| E["Volume stays constant<br/>V₁ = V₂"]
    C -->|"Cube (side a)"| C1["6a²"]
    C -->|"Cuboid (l,b,h)"| C2["2(lb + bh + hl)"]
    C -->|"Cylinder (r,h)"| C3["2πr(r + h)"]
    C -->|"Cone (r,l,h)"| C4["πr(r + l)"]
    C -->|"Sphere (r)"| C5["4πr²"]
    D -->|"Cube"| D1["a³"]
    D -->|"Cuboid"| D2["l × b × h"]
    D -->|"Cylinder"| D3["πr²h"]
    D -->|"Cone"| D4["⅓πr²h"]
    D -->|"Sphere"| D5["⁴⁄₃πr³"]

Alternative Method — Formula Sheet Comparison

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

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

Remember the volume ratio: Cone : Hemisphere : Cylinder = $1 : 2 : 3$ (for same base radius and height = radius). This is tested directly in CBSE 10. Archimedes was so proud of discovering this that he had it engraved on his tombstone.

Common Mistake

In cone problems, students confuse height ( $h$ ) with slant height ( $l$ ). The volume formula uses $h$ (vertical height), while the curved surface area uses $l$ (slant height). They’re related by $l = \sqrt{r^2 + h^2}$ . Using $l$ in the volume formula or $h$ in the CSA formula gives wrong answers. Always check which one the question provides.

Question

Solution — Step by Step

Why This Works

Alternative Method — Formula Sheet Comparison

Common Mistake

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