V = (πh/3)(r₁² + r₂² + r₁r₂). Cutting a cone parallel to base.

Frustum of Cone — Volume and Surface Area Formulas

Question

A solid cone of height 24 cm and base radius 6 cm is cut by a plane parallel to its base at a height of 8 cm from the base. Find the volume and total surface area of the frustum (the lower portion).

This is a direct application question from CBSE 2025 Sample Paper — frustums appear almost every year in board exams with exactly this setup.

Solution — Step by Step

The original cone has height $H = 24$ cm and base radius $R = 6$ cm. We cut at height $h_{\text{cut}} = 8$ cm from the base, so the frustum height is $h = 8$ cm.

We need the radius of the top face ( $r$ ) of the frustum — this is the radius of the smaller cone that was removed.

Here’s the key idea: the cut creates a smaller cone on top that is similar to the original cone. The smaller cone has height $= 24 - 8 = 16$ cm.

By similarity of triangles (the cone’s cross-section is a triangle):

\frac{r}{R} = \frac{h_{\text{small}}}{H} = \frac{16}{24} = \frac{2}{3}

So $r = \frac{2}{3} \times 6 = 4$ cm.

We need slant height $l$ for surface area. The slant height connects the rim of the bottom circle to the rim of the top circle:

l = \sqrt{h^2 + (R - r)^2} = \sqrt{8^2 + (6 - 4)^2} = \sqrt{64 + 4} = \sqrt{68} = 2\sqrt{17} \text{ cm}

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

V = \frac{\pi \times 8}{3}(6^2 + 4^2 + 6 \times 4)

= \frac{8\pi}{3}(36 + 16 + 24) = \frac{8\pi}{3} \times 76 = \frac{608\pi}{3} \approx 636.9 \text{ cm}^3

\text{TSA} = \pi l(R + r) + \pi R^2 + \pi r^2

= \pi \cdot 2\sqrt{17}(6 + 4) + \pi(36) + \pi(16)

= 20\sqrt{17}\,\pi + 52\pi = \pi(20\sqrt{17} + 52)

\approx \pi(82.46 + 52) \approx \pi \times 134.46 \approx 422.6 \text{ cm}^2

Final Answers:

Volume $= \dfrac{608\pi}{3}$ cm³ $\approx 636.9$ cm³
Total Surface Area $= \pi(20\sqrt{17} + 52)$ cm² $\approx 422.6$ cm²

Why This Works

The frustum formula $V = \frac{\pi h}{3}(R^2 + r^2 + Rr)$ comes directly from subtracting the volume of the smaller (removed) cone from the original cone. When you expand and simplify, this compact form appears — it’s elegant, and worth deriving once so it sticks.

The slant height $l = \sqrt{h^2 + (R-r)^2}$ is the Pythagoras theorem applied to the trapezoid cross-section. The “base” of the right triangle is $(R - r)$ , the perpendicular rise is $h$ , and the hypotenuse is $l$ .

The similar-triangles step for finding $r$ is where the real thinking happens in this problem. Once you have $r$ , the rest is formula application.

Alternative Method

You can find volume by the subtraction method — no formula memorisation needed.

Volume of original cone $= \frac{1}{3}\pi R^2 H = \frac{1}{3}\pi(36)(24) = 288\pi$ cm³

Volume of removed smaller cone $= \frac{1}{3}\pi r^2 h_{\text{small}} = \frac{1}{3}\pi(16)(16) = \frac{256\pi}{3}$ cm³

Frustum volume $= 288\pi - \frac{256\pi}{3} = \frac{864\pi - 256\pi}{3} = \frac{608\pi}{3}$ cm³ ✓

This confirms our answer. For PYQs where you might blank on the frustum formula, this method always works.

In CBSE marking schemes, leaving the answer in terms of $\pi$ is accepted and often preferred. You lose marks for arithmetic errors when you substitute $\pi = 3.14$ and calculate wrong. Keep it exact unless the question specifically says “use $\pi = 22/7$ ”.

Common Mistake

Using the wrong height for the smaller cone. Students often take the smaller cone’s height as 8 cm (the cut height from base) instead of 16 cm (the height from the cut to the apex). The frustum is the bottom portion — the smaller cone is the top portion, which has height $24 - 8 = 16$ cm. Using 8 cm gives $r = 2$ cm, which leads to a completely wrong surface area. Always draw a rough diagram and mark the apex at the top.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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