Wentworth's Plane and Solid Geometry · Book VIII, Exs. 836–865 · Book 9: Book VIII: The Sphere

132. Miscellaneous exercises (2)

Goal: Your student finishes solid geometry with constructions, cylinders, cones and spheres, and the famous sphere-and-cylinder result.

⏱ About 45 minutes · 6 steps

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You'll need

  • The book, open to these pages
  • A ruler (straightedge) and a pair of compasses
  • Sharp pencil, eraser and plain paper (or a geometry notebook)
  • A calculator (optional)

What's in the book

Pages 407–408: Miscellaneous Exercises, Ex. 836–865. This ends Book VIII and the solid geometry proper.

The lesson, step by step

  1. Warm up

    Ask: Volume and surface of a sphere and of a cylinder of radius R and height 2R?

  2. The sphere and the cylinder

    Ex. 857. The volume of a sphere is two thirds the volume of the circumscribed circular cylinder, and its surface is two thirds the total surface of the cylinder.

    Work it out with the formulas. Many people regard this as one of the most beautiful facts in geometry.

  3. Computations

    Ex. 844. Find the area of a spherical pentagon whose angles are 122°, 128°, 131°, 160°, 161°, if the surface of the sphere is 150 square feet.

    Ex. 859. A sphere 12 inches in diameter has an auger hole 3 inches in diameter through its centre. Find the remaining volume.

  4. Turning figures

    Ex. 861. Compare the volumes of the solids generated by a rectangle turning successively about two adjacent sides, the lengths of these sides being a and b.

    Ex. 863. An equilateral triangle revolves about one of its altitudes. Compare the volumes of the solids generated by the triangle, the inscribed circle, and the circumscribed circle.

  5. Final review

    Ask your student to choose the five most useful formulas of solid geometry and explain where each comes from.

  6. Wrap up

    Solid geometry is finished. Book IX, on the conic sections, follows.

Answers to the book's problems

Ex. 839: πR²H = 2πRH × ½R. Ex. 841: H = 2R: V = 2πR³ and T = 6πR², and 6πR² × ⅓R = 2πR³. Ex. 843: R/2. Ex. 844: the angles total 702°, so E = 702 − 540 = 162, and the area is 150 × 162/720 = 33.75 sq ft. Ex. 854: R ≈ 1.86 ft, surface ≈ 43.5 sq ft. Ex. 857: sphere 4/3 πR³ and 4πR²; cylinder 2πR³ and 6πR². Ex. 859: the hole leaves a ring of height h = 2√(36 − 2.25) ≈ 11.62 in, whose volume is πh³/6 ≈ 821 cu in. Ex. 860: two cones: πa²√3. Ex. 861: b : a (about side a: πb²a; about side b: πa²b). Ex. 862: cone ½πa², sphere ⅓πa²: 3 : 2. Ex. 863: 9 : 4 : 32.

Worked out for this site from the scan. If a number in your copy differs, trust the book.

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