Wentworth's Plane and Solid Geometry · Book VIII, Props. XXVIII–XXX · Book 9: Book VIII: The Sphere
127. Areas of lunes, spherical triangles and polygons
Goal: Your student can find the area of a lune, and of a spherical triangle or polygon from its spherical excess.
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 ball (an orange or a rubber ball) and a rubber band or string
- A calculator (optional)
What's in the book
Propositions XXVIII–XXX (§§ 829–837) with corollaries, and Exercises 759–767.
The lesson, step by step
Warm up
Ask: What is the spherical excess of a triangle? How many spherical degrees cover a sphere? (720.)
The new idea
The area of a lune is to the area of the surface of the sphere as the number of degrees in the angle of the lune is to 360.
The area of a spherical triangle, expressed in spherical degrees, is numerically equal to the spherical excess of the triangle.
§ 836: area of a spherical triangle = πR² × E ÷ 180.
Prop. XXX: for a polygon of n sides with angle sum T, the area in spherical degrees is T − (n − 2) 180.
Draw the figure
On the ball, extend the three sides of a triangle into full great circles. They cut the surface into eight triangles, and three lunes each contain the triangle.
Walk through the proof
XXVIII: divide the circumference into equal parts and pass great circles through them; the lunes are as their angles.
XXIX: triangle ABC with each neighbor forms a lune of angle A, B or C. Using symmetrical triangles, the three lunes cover a hemisphere (360 spherical degrees) plus the triangle twice more. So 2(A + B + C) = 360 + 2 × triangle, and triangle = A + B + C − 180.
XXX: split the polygon into n − 2 triangles from one vertex.
Your student proves it back
Close the book. Your student sketches the figure from memory (a simple perspective drawing is fine), letters it the same way, and gives the proof aloud: what is given, what is to be proved, then each step with its reason ("by § 501", "by § 507"). Point to each line and plane as it is named.
If a step is missing, ask "How do we know that?" rather than supplying it.
Exercises
Ex. 759. Given the radius of a sphere 10 inches. Find the area of a lune whose angle is 30°.
Ex. 761. What part of the surface of a sphere is a triangle whose angles are 120°, 100°, and 95°? What is its area in square inches, if the radius of the sphere is 6 inches?
Ex. 763. If the radii of two spheres are 6 inches and 4 inches respectively, and the distance between their centres is 5 inches, what is the area of the circle of intersection of these spheres?
Wrap up
Ask: What is the area of a tri-rectangular triangle on a sphere of radius R? (E = 90, so ½πR², one eighth of the sphere.)
Answers to the book's problems
Ex. 759: πR²A/90 = 100π/3 ≈ 104.7 sq in. Ex. 760: use the same formula with R = 8 in. Ex. 761: E = 135, so 135/720 = 3/16 of the surface; 27π ≈ 84.8 sq in. Ex. 762: E = 180, R = 8: 64π ≈ 201.1 sq in. Ex. 763: the plane of the circle is 4.5 in from the larger centre; radius² = 36 − 20.25 = 15.75; area ≈ 49.5 sq in. Ex. 764: √5.25 ≈ 2.29 in. Ex. 765: π(R'² − R²). Ex. 766: a circle of radius √18.75 ≈ 4.33 in, whose plane is perpendicular to AB at 2.5 in from A. Ex. 767: if the sections are on the same side of the centre, R² = a² + x² = b² + (x + d)² for the distance x; solve for x, then R.
Worked out for this site from the scan. If a number in your copy differs, trust the book.