What is a Truncated Cone?

A truncated cone, also known as a frustum of a cone, is a three-dimensional geometric shape that results from slicing the top off a cone parallel to its base. This creates two circular faces: a smaller top face and a larger bottom face.

Key Properties of a Truncated Cone

Bases and Height

  • Bases: The two circular faces are called the bases. The smaller circular face is known as the upper base, and the larger circular face is the lower base.
  • Height: The perpendicular distance between the two bases is the height (h) of the truncated cone.

Slant Height

The slant height (s) is the distance along the lateral surface between the edges of the two bases. It can be calculated using the Pythagorean theorem if the height and the radii of the bases are known.

Radii

  • Upper Radius: The radius of the smaller, upper base is denoted as $r_1$
  • Lower Radius: The radius of the larger, lower base is denoted as $r_2$

Formulas for Surface Area and Volume

Surface Area

The surface area of a truncated cone is the sum of the areas of the two bases and the lateral surface area.

Formula

$A = text{Base Area}_1 + text{Base Area}_2 + text{Lateral Area}$
$A = pi r_1^2 + pi r_2^2 + pi (r_1 + r_2) s$

Volume

The volume of a truncated cone can be calculated using the formula:

Formula

$V = frac{1}{3} pi h (r_1^2 + r_2^2 + r_1 r_2)$

Example Calculation

Let’s say we have a truncated cone with the following measurements:

  • Upper radius ($r_1$) = 3 units
  • Lower radius ($r_2$) = 5 units
  • Height (h) = 4 units

Surface Area Calculation

First, we need to find the slant height (s). Using the Pythagorean theorem:
$s = sqrt{(r_2 – r_1)^2 + h^2}$
$s = sqrt{(5 – 3)^2 + 4^2}$
$s = sqrt{4 + 16}$
$s = sqrt{20}$
$s = 2sqrt{5}$

Now, we can calculate the surface area:
$A = pi (3^2) + pi (5^2) + pi (3 + 5) 2sqrt{5}$
$A = 9pi + 25pi + 16pisqrt{5}$
$A = 34pi + 16pisqrt{5}$

Volume Calculation

$V = frac{1}{3} pi (4) (3^2 + 5^2 + 3 cdot 5)$
$V = frac{1}{3} pi (4) (9 + 25 + 15)$
$V = frac{1}{3} pi (4) (49)$
$V = frac{1}{3} pi (196)$
$V = 65.33 pi$

Conclusion

Understanding a truncated cone’s properties and formulas is essential for solving real-world problems involving this shape, such as calculating the volume of a truncated cone-shaped container or the surface area for material requirements.

Citations

  1. 1. Khan Academy – Geometry
  2. 2. Math is Fun – Truncated Cone
  3. 3. Wolfram MathWorld – Frustum

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ