Right Circular Cone

A right circular cone is the 3D shape you get by rotating a right triangle about one of its legs , or the shape of a party hat or an ice-cream cone. A single circular base, a single vertex, and a curved surface joining them. This lesson defines the slant height and gives all the formulas.

Definitions

A right circular cone has a circular base of radius $r$ and a vertex (apex) directly above the centre of the base at height $h$ , the perpendicular distance from vertex to base.

The slant height $\ell$ is the distance from the vertex to any point on the circumference of the base. By the Pythagoras theorem in the right triangle formed (radius, height, slant height): $\ell = \sqrt{r^2 + h^2}.$

Formulas

For a cone of radius $r$, height $h$, slant height $\ell$:

• Curved surface area (CSA): $\pi r \ell$.
• Total surface area (TSA): $\pi r \ell + \pi r^2 = \pi r(\ell + r)$.
• Volume: $V = \tfrac{1}{3} \pi r^2 h$.

Why the formulas are right

Curved surface. Cut the curved surface vertically and unroll it. You get a sector of a circle with radius $\ell$ and arc length $2\pi r$ (the base circumference). The area of this sector is $\tfrac{1}{2} \times \text{radius} \times \text{arc length} = \tfrac{1}{2} \ell \cdot 2\pi r = \pi r \ell$.

Total surface. Add the base circle of area $\pi r^2$. Total: $\pi r \ell + \pi r^2 = \pi r(\ell + r)$.

Volume. This is one-third the volume of the cylinder with the same base radius and height: $V = \tfrac{1}{3} \pi r^2 h$. The $\tfrac{1}{3}$ factor can be proved by calculus or by experiment (filling three cones with water fills exactly one cylinder of the same base and height). We accept it here.

The cylinder-cone relationship

If a cylinder and a cone have the same radius and the same height:

• Cylinder volume $= \pi r^2 h$.
• Cone volume $= \tfrac{1}{3} \pi r^2 h$.

So cone volume is one-third of cylinder volume. This is a beautiful and remarkable fact.

Worked examples

Example 1. A cone has radius $7$ cm and height $24$ cm. Find slant height, CSA, TSA, V.

$\ell = \sqrt{49 + 576} = \sqrt{625} = 25$ cm. $\text{CSA} = \pi r \ell = \tfrac{22}{7} \cdot 7 \cdot 25 = 550$ cm$^2$. $\text{TSA} = \pi r(\ell + r) = \tfrac{22}{7} \cdot 7 \cdot 32 = 704$ cm$^2$. $V = \tfrac{1}{3} \pi r^2 h = \tfrac{1}{3} \cdot \tfrac{22}{7} \cdot 49 \cdot 24 = 1232$ cm$^3$.

Example 2. A conical tent has radius $10$ m and height $24$ m. Find the canvas needed (CSA).

$\ell = \sqrt{100 + 576} = 26$ m. CSA $= \pi r \ell = \tfrac{22}{7} \cdot 10 \cdot 26 \approx 817.14$ m$^2$.

Example 3. A conical heap of sand has base radius $3$ m and height $4$ m. Find the volume.

$V = \tfrac{1}{3} \pi r^2 h = \tfrac{1}{3} \cdot \tfrac{22}{7} \cdot 9 \cdot 4 = \tfrac{264}{7} \approx 37.71$ m$^3$.

Example 4. Find the slant height of a cone with radius $5$ and height $12$.

$\ell = \sqrt{25 + 144} = \sqrt{169} = 13$.

Example 5. A cone has CSA $440$ cm$^2$ and slant height $20$ cm. Find the radius. (Use $\pi = \tfrac{22}{7}$.)

$\pi r \ell = 440 \Rightarrow \tfrac{22}{7} \cdot r \cdot 20 = 440 \Rightarrow r = \tfrac{440 \cdot 7}{22 \cdot 20} = 7$ cm.

Try it yourself

1. Cone: $r = 6, h = 8$. Find slant height, CSA, TSA, V.
2. A conical hat has radius $7$ and height $24$. How much paper is needed (CSA only)?
3. A conical heap of sand has radius $3$ and height $4$. Find the volume.
4. CSA of a cone is $308$ cm$^2$, radius $7$. Find the slant height.
5. A right triangle with legs $3$ and $4$ is rotated about the side of length $3$. Find the volume of the cone formed.
6. Compare the volumes of a cylinder of radius $5$ and height $10$ with a cone of the same radius and height.
7. Find $h$ if a cone has $r = 7, \ell = 25$.
8. A cone of radius $6$ has volume $\pi \cdot 6^2 \cdot 4$. Find the height. (Be careful , what's the formula?)
9. A cone has volume $1232$ cm$^3$ and radius $7$. Find the height ($\pi = \tfrac{22}{7}$).
10. Two cones have the same height; the second has twice the radius. Compare their volumes.

Pitfalls / Insight

• Slant height $\ell$ vs. perpendicular height $h$. Use $\ell$ in CSA formula; use $h$ in volume formula.
• Don't forget the $\tfrac{1}{3}$. Cone volume is $\tfrac{1}{3} \pi r^2 h$, not $\pi r^2 h$.
• The base is a circle of area $\pi r^2$. TSA adds this to the CSA.

Insight. A cone is one-third of its containing cylinder. This single fact ties together all 3D geometry , the cone, cylinder, and sphere are all linked by similar fractional relationships in volume.