Solid Geometry

LessonAbout 1.5 h2 of 2 in Plane and Solid Geometry

Key Ideas

  • Solid geometry measures three-dimensional space and volumes.
  • Polyhedrons are solids bounded by plane polygonal faces.
  • Prisms, pyramids, cylinders, and cones form foundational solids.
  • Spherical geometry examines solids bounded by spherical surfaces.

Solid Geometry is the branch of geometry that deals with three-dimensional figures, their surface areas, volumes, and geometric properties in space.

Polyhedrons and Platonic Solids

  • Regular Polyhedron: solid with identical regular polygon faces.
  • Five Platonic Solids: tetrahedron, hexahedron, octahedron, dodecahedron, icosahedron.
  • Euler’s Polyhedron Formula: relates faces, vertices, and edges (\(F + V = E + 2\)).
  • Cube: regular hexahedron with 6 square faces (\(V = a^3\)).
  • Rectangular Parallelepiped: solid with 6 rectangular faces (\(V = abc\)).

Volume of Regular Tetrahedron Formula

\[ V = \dfrac{a^3}{6\sqrt{2}} \]

Volume of Regular Octahedron Formula

\[ V = \dfrac{a^3\sqrt{2}}{3} \]

Example 1: Calculate the total volume of a regular tetrahedron having an edge length of 6 cm.

Solution: Apply the regular tetrahedron volume formula for \(a = 6\text{ cm}\):

\[ V = \dfrac{6^3}{6\sqrt{2}} = \dfrac{216}{6\sqrt{2}} = \dfrac{36}{\sqrt{2}} = 25.46\text{ cm}^3 \]

Answer: 25.46 cm³

Example 2: A regular octahedron has an edge length of 2 m. Find its volume.

Solution: Apply the regular octahedron volume formula for \(a = 2\text{ m}\):

\[ V = \dfrac{(2)^3\sqrt{2}}{3} = \dfrac{8\sqrt{2}}{3} = 3.77\text{ m}^3 \]

Answer: 3.77 m³

Example 3: The space diagonal of a cube is \(4\sqrt{3}\text{ m}\). Find its total surface area and volume.

Solution: Let \(a\) be the edge of the cube. The space diagonal is \(d = a\sqrt{3}\):

\[ a\sqrt{3} = 4\sqrt{3} \implies a = 4\text{ m} \]
\[ A_s = 6a^2 = 6(4)^2 = 96\text{ m}^2 \]
\[ V = a^3 = (4)^3 = 64\text{ m}^3 \]

Answer: Surface area = 96 m², Volume = 64 m³

Prisms and Cylinders

  • Prism Volume: product of base area and altitude (\(V = A_b h\)).
  • Right Prism: lateral faces are perpendicular to the bases.
  • Oblique Prism: lateral faces are inclined to the bases (\(V = A_R L\)).
  • Truncated Triangular Prism: prism with non-parallel truncated base (\(V = A_R \cdot \dfrac{h_1 + h_2 + h_3}{3}\)).
  • Cylinder Volume: circular base area times height (\(V = \pi r^2 h\)).

Prism Volume Formula

\[ V = A_b h = A_R L \]

Truncated Triangular Prism Volume Formula

\[ V = A_{\text{base}} \cdot \left(\dfrac{h_1 + h_2 + h_3}{3}\right) \]

Example 4: The base of a right prism is a regular hexagon with side length 6 cm. If the bases are 12 cm apart, determine the volume of the prism.

Solution: Compute the area of the regular hexagonal base \(A_b\):

\[ A_b = \dfrac{6(6)^2}{4 \tan 30^\circ} = \dfrac{216}{4(0.57735)} = 93.53\text{ cm}^2 \]
\[ V = A_b h = (93.53)(12) = 1122.37\text{ cm}^3 \]

Answer: 1122.37 cm³

Example 5: Determine the volume of a right truncated triangular prism whose base has side lengths 10 ft, 9 ft, and 12 ft, and whose perpendicular heights at the vertices are 8.6 ft, 7.1 ft, and 5.5 ft.

Solution: Find base area \(A_{\text{base}}\) using Heron’s formula with \(s = \dfrac{10 + 9 + 12}{2} = 15.5\text{ ft}\):

\[ A_{\text{base}} = \sqrt{15.5(15.5 - 10)(15.5 - 9)(15.5 - 12)} = \sqrt{15.5 \times 5.5 \times 6.5 \times 3.5} = 44.04\text{ ft}^2 \]
\[ h_{\text{mean}} = \dfrac{8.6 + 7.1 + 5.5}{3} = \dfrac{21.2}{3} = 7.067\text{ ft} \]
\[ V = A_{\text{base}} \cdot h_{\text{mean}} = (44.04)(7.067) = 311.2\text{ ft}^3 \]

Answer: 311.2 ft³

Example 6: A right circular cylinder with a volume of 6.54 m³ is circumscribed about a right prism with an equilateral triangular base of side 1.25 m. Find the altitude of the cylinder.

Solution: The radius \(r\) of the circumscribed circle around an equilateral triangle of side \(x = 1.25\text{ m}\) is:

\[ r = \dfrac{x}{\sqrt{3}} = \dfrac{1.25}{1.732} = 0.7217\text{ m} \]

Solve for altitude \(h\) from cylinder volume \(V = \pi r^2 h\):

\[ 6.54 = \pi (0.7217)^2 h = 1.636 h \implies h = 4.00\text{ m} \]

Answer: 4.00 m

Pyramids, Cones, and Frustums

  • Pyramid/Cone Volume: one-third base area times height (\(V = \dfrac{1}{3} A_b h\)).
  • Right Circular Cone Lateral Area: product of \(\pi\), radius, and slant height (\(A_L = \pi r L\)).
  • Frustum of Pyramid/Cone: portion between base and parallel cutting plane.
  • Frustum Volume Formula: incorporates upper and lower base areas.

Pyramid and Cone Volume Formula

\[ V = \dfrac{1}{3} A_b h \]

Frustum Volume Formula

\[ V = \dfrac{h}{3} \left(A_1 + A_2 + \sqrt{A_1 A_2}\right) \]

Example 7: A regular triangular pyramid has an altitude of 9 m and a volume of 187.06 m³. Find the length of the base edge.

Solution: Let \(x\) be the base edge of the equilateral triangular base:

\[ A_b = \dfrac{1}{2} x^2 \sin 60^\circ = 0.433 x^2 \]
\[ V = \dfrac{1}{3} A_b h \implies 187.06 = \dfrac{1}{3} (0.433 x^2) (9) \]
\[ 187.06 = 1.299 x^2 \implies x^2 = 144 \implies x = 12\text{ m} \]

Answer: 12 m

Example 8: A frustum of a cone has a lower base diameter of 15 cm, an upper base diameter of 10 cm, and an altitude of 25 cm. Find its volume.

Solution: Lower base radius \(R = 7.5\text{ cm}\) (\(A_1 = 56.25\pi\)), upper base radius \(r = 5\text{ cm}\) (\(A_2 = 25\pi\)):

\[ V = \dfrac{\pi h}{3} \left(R^2 + r^2 + Rr\right) = \dfrac{25\pi}{3} \left(7.5^2 + 5^2 + (7.5)(5)\right) \]
\[ V = \dfrac{25\pi}{3} (56.25 + 25 + 37.5) = \dfrac{25\pi}{3} (118.75) = 3108.87\text{ cm}^3 \]

Answer: 3108.87 cm³

Example 9: A circular cardboard with a diameter of 1 m is made into a conical hat 40 cm high by cutting off a sector and joining the edges. Find the central angle of the sector removed.

Solution: Cardboard radius \(r_{\text{sheet}} = 50\text{ cm}\) becomes cone slant height \(L = 50\text{ cm}\). Cone height \(h = 40\text{ cm}\). Base radius of cone \(x\):

\[ x = \sqrt{L^2 - h^2} = \sqrt{50^2 - 40^2} = 30\text{ cm} \]

Arc length of sector removed equals sheet circumference minus cone base circumference:

\[ C = 2\pi(50) - 2\pi(30) = 40\pi\text{ cm} \]

Angle subtended by removed sector \(\theta\):

\[ \theta = \dfrac{C}{r_{\text{sheet}}} = \dfrac{40\pi}{50} = 0.8\pi\text{ rad} = 0.8 \times 180^\circ = 144^\circ \]

Answer: 144°

Spheres and Spherical Solids

  • Sphere Volume and Area: \(V = \dfrac{4}{3}\pi R^3\) and \(A = 4\pi R^2\).
  • Zone Area: spherical surface bounded by parallel planes (\(A = 2\pi R h\)).
  • Spherical Segment: solid bounded by a zone and parallel base planes.
  • Spherical Sector: solid generated by rotating a circle sector about an axis.
  • Spherical Wedge: portion bounded by two half great circles (\(V = \dfrac{\pi R^3 \theta^\circ}{270^\circ}\)).
  • Spherical Pyramid: pyramid formed by a spherical polygon base and center vertex.

Sphere Surface Area and Volume Formulas

\[ A = 4\pi R^2, \qquad\qquad V = \dfrac{4}{3}\pi R^3 \]

Zone Area Formula

\[ A_{\text{zone}} = 2\pi R h \]

Spherical Segment Volume Formula

\[ V = \dfrac{\pi h^2}{3} (3R - h) \]

Spherical Wedge Volume Formula

\[ V = \dfrac{\pi R^3 \theta^\circ}{270^\circ} = \dfrac{2}{3} R^3 \theta_{\text{rad}} \]

Spherical Pyramid Volume Formula

\[ V = \dfrac{\pi R^3 E}{540^\circ} \]

Example 10: A sphere having a diameter of 30 cm is cut into 2 segments. If the altitude of the first segment is 6 cm, find the ratio of the surface area of the second segment to that of the first.

Solution: Sphere radius \(R = 15\text{ cm}\). Total altitude \(2R = 30\text{ cm}\). Altitudes of segments are \(h_1 = 6\text{ cm}\) and \(h_2 = 30 - 6 = 24\text{ cm}\). Zone area is proportional to altitude (\(A = 2\pi R h\)):

\[ \dfrac{A_2}{A_1} = \dfrac{2\pi R h_2}{2\pi R h_1} = \dfrac{h_2}{h_1} = \dfrac{24}{6} = 4 \]

Answer: 4:1

Example 11: Determine the volume of a spherical segment of height 6 m in a sphere of diameter 30 m, and find its zone area.

Solution: Sphere radius \(R = 15\text{ m}\), segment altitude \(h = 6\text{ m}\):

\[ V = \dfrac{\pi(6)^2}{3} \left(3(15) - 6\right) = 12\pi (45 - 6) = 12\pi (39) = 1470.265\text{ m}^3 \]
\[ A_{\text{zone}} = 2\pi R h = 2\pi (15) (6) = 180\pi = 565.5\text{ m}^2 \]

Answer: Volume = 1470.265 m³, Zone area = 565.5 m²

Example 12: Find the volume of a spherical wedge of radius 1 unit if its central angle is 1 radian.

Solution: Apply spherical wedge volume formula for \(R = 1\) and \(\theta = 1\text{ rad}\):

\[ V = \dfrac{2}{3} R^3 \theta = \dfrac{2}{3} (1)^3 (1) = \dfrac{2}{3}\text{ cubic units} \]

Answer: 2/3 cubic units

Prismatoids, Tori, and Solids of Revolution

  • Prismatoid: polyhedron with parallel polygonal bases and trapezoidal/triangular lateral faces.
  • Prismoidal Formula: exact volume formula for prismoidal solids (\(V = \dfrac{L}{6}(A_1 + 4A_m + A_2)\)).
  • Torus: donut-shaped solid formed by revolving a circle about an external line.
  • Ellipsoid / Spheroid: solid formed by revolving an ellipse (\(V = \dfrac{4}{3}\pi a b c\)).
  • Pappus Second Proposition: volume equals generating area times distance traveled by its centroid (\(V = 2\pi \bar{x} A\)).

Prismoidal Formula

\[ V = \dfrac{L}{6} \left(A_1 + 4A_m + A_2\right) \]

Torus Volume and Surface Area Formulas

\[ V = 2\pi^2 R r^2, \qquad\qquad A = 4\pi^2 R r \]

Pappus Second Proposition (Volume of Revolution)

\[ V = 2\pi \bar{x} A \]

Example 13: Along a 200 m section of an excavation, cross-sectional areas taken every 50 m are 2556 m², 2619 m², 2700 m², 2610 m², and 2484 m². Calculate the total volume of excavation.

Solution: The 5 cross-sections form two consecutive 100 m prismoidal segments. Apply Prismoidal formula to each segment (\(L = 100\text{ m}\)):

\[ V_1 = \dfrac{100}{6} \left(2556 + 4(2619) + 2700\right) = \dfrac{100}{6} (15732) = 262,200\text{ m}^3 \]
\[ V_2 = \dfrac{100}{6} \left(2700 + 4(2610) + 2484\right) = \dfrac{100}{6} (15624) = 260,400\text{ m}^3 \]
\[ V_{\text{total}} = 262,200 + 260,400 = 522,600\text{ m}^3 \]

Answer: 522,600 m³

Example 14: Find the volume generated by revolving the ellipse \(\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1\) about the vertical line \(x = 3\).

Solution: Ellipse area \(A = \pi a b = \pi (3) (2) = 6\pi\). The centroid of the centered ellipse is at \((0,0)\). Distance from centroid to axis of rotation \(x = 3\) is \(\bar{x} = 3\). Apply Pappus second proposition:

\[ V = 2\pi \bar{x} A = 2\pi (3) (6\pi) = 36\pi^2 = 355.3\text{ cubic units} \]

Answer: 355.3 cubic units

Example 15: A solid has a circular base of radius \(r\). Every cutting plane perpendicular to a given diameter is a square. Determine the volume of the solid.

Solution: This solid is a prismatoid extending along diameter \(2r\) (\(L = 2r\)). End areas \(A_1 = 0\), \(A_2 = 0\). At mid-section, the chord length is \(2r\), so mid-section square area \(A_m = (2r)^2 = 4r^2\). Apply prismoidal formula:

\[ V = \dfrac{2r}{6} \left(0 + 4(4r^2) + 0\right) = \dfrac{2r}{6} (16r^2) = \dfrac{16}{3} r^3 \]

Answer: 16r³/3

Formula Summary

Formula Name Formula
Volume of Regular Tetrahedron Formula \(V = \dfrac{a^3}{6\sqrt{2}}\)
Volume of Regular Octahedron Formula \(V = \dfrac{a^3\sqrt{2}}{3}\)
Prism Volume Formula \(V = A_b h = A_R L\)
Truncated Triangular Prism Volume Formula \(V = A_{\text{base}} \cdot \left(\dfrac{h_1 + h_2 + h_3}{3}\right)\)
Pyramid and Cone Volume Formula \(V = \dfrac{1}{3} A_b h\)
Frustum Volume Formula \(V = \dfrac{h}{3} \left(A_1 + A_2 + \sqrt{A_1 A_2}\right)\)
Sphere Surface Area and Volume Formulas \(A = 4\pi R^2, \quad V = \dfrac{4}{3}\pi R^3\)
Zone Area Formula \(A_{\text{zone}} = 2\pi R h\)
Spherical Segment Volume Formula \(V = \dfrac{\pi h^2}{3} (3R - h)\)
Spherical Wedge Volume Formula \(V = \dfrac{\pi R^3 \theta^\circ}{270^\circ} = \dfrac{2}{3} R^3 \theta_{\text{rad}}\)
Spherical Pyramid Volume Formula \(V = \dfrac{\pi R^3 E}{540^\circ}\)
Prismoidal Formula \(V = \dfrac{L}{6} \left(A_1 + 4A_m + A_2\right)\)
Torus Volume and Surface Area Formulas \(V = 2\pi^2 R r^2, \quad A = 4\pi^2 R r\)
Pappus Second Proposition (Volume of Revolution) \(V = 2\pi \bar{x} A\)