Volume of frustum — derivation and problem solving approach

Question

A bucket is in the shape of a frustum with top radius 15 cm, bottom radius 10 cm, and height 24 cm. Find its volume and the area of the metal sheet used to make it (CSA + bottom circle).

(CBSE Class 10 pattern)

Solution — Step by Step

flowchart TD
    A["Full Cone"] -->|"Cut horizontally\nparallel to base"| B["Small cone\n(removed)"]
    A -->|"What remains"| C["Frustum\n(bucket shape)"]
    C --> D["R = top radius\nr = bottom radius\nh = height"]
    D --> E["Volume = πh/3 × (R² + r² + Rr)"]
    D --> F["Slant height\nl = √(h² + (R-r)²)"]
    F --> G["CSA = π(R+r)l"]

V = \frac{\pi h}{3}(R^2 + r^2 + Rr)

= \frac{\pi \times 24}{3}(15^2 + 10^2 + 15 \times 10)

= 8\pi(225 + 100 + 150) = 8\pi \times 475 = 3800\pi

= \mathbf{11942.86 \text{ cm}^3 \approx 11.94 \text{ litres}}

l = \sqrt{h^2 + (R - r)^2} = \sqrt{24^2 + (15-10)^2} = \sqrt{576 + 25} = \sqrt{601} \approx 24.52 \text{ cm}

The bucket has a curved surface and a bottom circle (no top — it is open).

CSA of frustum: $\pi(R + r)l = \pi(15 + 10) \times 24.52 = 613\pi \approx 1925.22$ cm²

Bottom circle: $\pi r^2 = \pi \times 100 = 100\pi \approx 314.16$ cm²

Total metal sheet: $613\pi + 100\pi = 713\pi \approx \mathbf{2239.38 \text{ cm}^2}$

Why This Works

A frustum is simply a cone with the top sliced off. Its volume equals the volume of the full cone minus the volume of the small cone that was removed. When you work through this subtraction algebraically (using similar triangles to relate the heights and radii), the formula $V = \frac{\pi h}{3}(R^2 + r^2 + Rr)$ emerges.

The term $R^2 + r^2 + Rr$ is the key — it is not $(R+r)^2$ or $R^2 + r^2$ . The cross term $Rr$ comes from the subtraction of similar cones. Remembering this specific form is critical.

Alternative Method — Direct Derivation

If the full cone has height $H$ and the small removed cone has height $H - h$ :

By similar triangles: $\frac{r}{R} = \frac{H-h}{H}$ , so $H = \frac{Rh}{R-r}$ .

Volume of frustum = Volume of big cone - Volume of small cone = $\frac{\pi}{3}R^2 H - \frac{\pi}{3}r^2(H-h)$

Substituting and simplifying gives $\frac{\pi h}{3}(R^2 + r^2 + Rr)$ .

In CBSE, frustum problems are usually about practical objects: buckets, flower pots, or milk containers. The question will give two radii and a height (or slant height). Always identify which is $R$ (larger) and $r$ (smaller). If slant height $l$ is given instead of vertical height $h$ , use $h = \sqrt{l^2 - (R-r)^2}$ to convert.

Common Mistake

The most common error: writing the volume formula as $\frac{\pi h}{3}(R^2 + r^2)$ and forgetting the $Rr$ term. This underestimates the volume. The correct formula has three terms inside: $R^2 + r^2 + Rr$ . A quick sanity check: if $R = r$ (cylinder, not a frustum), the formula should give $\pi r^2 h$ , and indeed $\frac{\pi h}{3}(r^2 + r^2 + r^2) = \pi r^2 h$ . Without the $Rr$ term, this check fails.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Derivation

Common Mistake

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