Tenth Grade Mathematics: Frustum of a Cone: Slant Height, Surface Area, and Volume
When a right circular cone is sectioned by a cutting plane parallel to its base and the smaller upper cone containing the apex is discarded, the remaining lower solid is called the Frustum of a Cone. Found everywhere in daily life as drinking glasses, buckets, flower pots, and lamp shades, the frustum is bounded by two unequal parallel circular bases and a tapering curved lateral surface. In this chapter, we derive the universal formulas for the slant height, curved surface area, total surface area, and volume of a frustum by analyzing the similarity between the original cone and the removed cone.
Geometric Structure of a Frustum
Let a right circular cone of height H, slant height L, and base radius r1 be sliced by a plane parallel to the base at height h, removing a small cone of height h1, slant height l1, and base radius r2 (where r1 > r2):
Geometry of a Frustum of a Cone:
Apex V
/ / \ Small Removed Cone
/ \ (Height h1, Radius r2, Slant l1)
------+------+------ Cutting Plane Parallel to Base
/ |r2 \ Top Circular Face (Radius r2)
+-------+--------+
| | |
| | | Frustum of Cone
Slant | |h | (Height h, Slant Height l)
Height (l)| | |
+-------+--------+ Bottom Circular Base (Radius r1)
/ |r1 +---------+----------+
Parameters of the Frustum:
- Radius of larger circular base: r1
- Radius of smaller circular top: r2 (r1 > r2)
- Vertical height: h
- Slant height: l
Derivation of Slant Height l
In the vertical axial cross-section of the frustum, drop a perpendicular from the edge of the top face to the bottom base: This forms a small right-angled triangle with: Vertical leg = h Horizontal leg = r1 - r2 Hypotenuse = Slant height l
By the Pythagorean theorem:
Theorem (Slant Height of a Frustum):
l = sqrt( h^2 + (r1 - r2)^2 )
Derivation of Volume and Surface Area via Similarity
In the vertical cross-section, the small cone and large cone form similar right triangles:
Similarity Ratio:
h1 / H = r2 / r1 and l1 / L = r2 / r1
Notice: H = h1 + h, and L = l1 + l
From l1 / (l1 + l) = r2 / r1: r1 * l1 = r2 * l1 + r2 * l ===> l1 * (r1 - r2) = r2 * l ===> l1 = (r2 * l) / (r1 - r2). Total slant height L = l1 + l = (r1 * l) / (r1 - r2).
1. Curved Surface Area (CSA)
The curved surface area of the frustum is the curved area of the large cone minus the curved area of the small cone:
CSA(Frustum) = pi * r1 * L - pi * r2 * l1
= pi * [ r1 * (r1 * l / (r1 - r2)) - r2 * (r2 * l / (r1 - r2)) ]
= pi * l * [ (r1^2 - r2^2) / (r1 - r2) ]
Notice difference of squares: (r1^2 - r2^2) / (r1 - r2) = r1 + r2.
Theorem (Curved Surface Area of a Frustum):
CSA = pi * (r1 + r2) * l
where l = sqrt( h^2 + (r1 - r2)^2 ).
2. Total Surface Area (TSA)
The total surface area includes the curved surface plus the areas of both circular faces:
Theorem (Total Surface Area of a Frustum):
TSA = pi * (r1 + r2) * l + pi * r1^2 + pi * r2^2
3. Volume of a Frustum
The volume of the frustum is the volume of the large cone minus the volume of the small cone:
Volume(Frustum) = (1/3) * pi * r1^2 * H - (1/3) * pi * r2^2 * h1
Using the similarity substitutions for H and h1:
Theorem (Volume of a Frustum):
1
V = --- * pi * h * (r1^2 + r2^2 + r1 * r2)
3
Notice the structural resemblance to the cubic identity: a^3 - b^3 = (a - b)(a^2 + b^2 + ab).
Chapter Exercises
Exercise 1: A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass. (Use pi = 22/7).
Exercise 2: The slant height of a frustum of a cone is 4 cm and the perimeters (circumferences) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Exercise 3: An open metal bucket is in the shape of a frustum of a cone mounted on a hollow cylindrical base. If the radii of the two circular ends of the bucket are 28 cm and 7 cm, its vertical height is 45 cm, and slant height is 49.2 cm, find the total area of metallic sheet used to make the bucket.
Solutions and Explanations
Solution 1: The height of the frustum is h = 14 cm. The diameters are 4 cm and 2 cm, giving radii r1 = 4/2 = 2 cm and r2 = 2/2 = 1 cm. The capacity of the glass is its internal volume: V = (1/3) * pi * h * (r1^2 + r2^2 + r1r2). Substitute the given values: V = (1/3) * (22/7) * 14 * (2^2 + 1^2 + 21) = (1/3) * 22 * 2 * (4 + 1 + 2) = (44 / 3) * 7 = 308 / 3 = 102.67 cm^3. Thus, the capacity of the glass is 102.67 cm^3.
Solution 2: Let the radii of the circular ends be r1 and r2. The circumferences are given as C1 = 2pir1 = 18 cm, which gives pir1 = 9 cm; and C2 = 2pir2 = 6 cm, which gives pir2 = 3 cm. Adding these two: pir1 + pir2 = pi(r1 + r2) = 9 + 3 = 12 cm. The slant height is l = 4 cm. The formula for the curved surface area of a frustum is: CSA = pi * (r1 + r2) * l. Substituting pi(r1 + r2) = 12 cm and l = 4 cm: CSA = 12 * 4 = 48 cm^2. Thus, the curved surface area of the frustum is 48 cm^2.
Solution 3: The bucket is an open container (open at the larger top, closed at the smaller bottom). The radii are r1 = 28 cm (top) and r2 = 7 cm (bottom). The slant height is l = 49.2 cm. The metal sheet covers: first, the curved surface of the frustum; second, the flat circular bottom of radius r2 = 7 cm (the top is open to hold liquid!). Total Area of Metal Sheet = CSA of Frustum + Area of Bottom Circle = pi * (r1 + r2) * l + pi * r2^2 = (22/7) * (28 + 7) * 49.2 + (22/7) * 7^2 = (22/7) * 35 * 49.2 + 22 * 7 = 22 * 5 * 49.2 + 154 = 110 * 49.2 + 154 = 5412 + 154 = 5566 cm^2. Thus, 5566 square centimeters of metal sheet is used.