Hippocrates of Chios
is a four-sided figure with one pair of parallel opposite sides, called
its bases. (In a parallelogram, both pairs of opposite sides are parallel.)
We typically orient a trapezoid so that its bases are horizontal with the
longer base below the shorter one, whence the height of the trapezoid is
the distance between the bases. Draw a trapezoid, label its longer base
its shorter base b, and its height h. By dissecting the trapezoid
into smaller figures, derive the formula
Evaluate these integrals
by sketching graphs of the corresponding regions:
The "lune on the triangle"
has an area calculation which is revealed by an argument similar to Hippocrates'
for the "lune on the square". Inscribe an equilateral triangle ABC
in a circle with center O and build semicircles on the sides of
the triangles. This creates three lunes, numbered 1, 2, 3 in the diagram
below. Clearly, the three lunes have the same area. Also, let D
be opposite end of the diameter of the large circle through B.
(a) A fundamental
result in geometry says that an angle of measure a
vertex lies on a circle cuts off an arc of the circle whose measure from
the center is 2a.
For instance, since angle BAD cuts off an arc whose measure from
the center of the circle is 180 degrees (namely arc BD), angle BAD
measures 90 degrees. What are the measures of the other two angles in triangle
BAD? What are the measures of the angles in triangle OAD?
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(b) It follows that
= (1/2)BD. Why?
(c) Let x
= AB be the length of the side of the equilateral triangle ABC,
and let d = BD be the diameter of the large circle. Then part (b)
says that AD = d/2. We can apply the Pythagorean Theorem to the
right triangle BAD. Show from this that
(d) It follows that the
ratio of the areas of the semicircle on diameter AB to the semicircle
on diameter BD is 3/4. Why?
(e) Use the diagram
to argue that, in terms of area, 3(semicircle on AB) + (triangle
= 3(lune) + 2(semicircle on BD)
and from this show
that the area of any one of the numbered lunes = (1/3)(triangle ABC)
+ (1/12)(semicircle BD) .
Hippocrates is said to
have attempted the quadrature of another lune, the "lune on the hexagon",
which we repeat here. Inscribe a regular hexagon (regular means
that all the sides and interior angles are equal) in a circle centered
at O. Drawing in the diagonals that join opposite vertices of the
hexagon dissects the hexagon into six triangles at the center C
of the circle. On each side of the hexagon build a semicircle. This constructs
six lunes, marked 1 through 6 in the diagram.
(a) In triangle ABC,
the angle at C is 60 degrees and sides AC = BC. Why?
(b) It follows that
the angles at A and B are equal. What do they measure? Use
this to explain why all six triangles are equilateral.
(c) The semicircles
containing lunes 1, 2, 3, and 4 have area equal to the upper half of the
large circle. Why?
(d) The three lunes
1, 2, and 3, together with the semicircle containing lune 4 have total
area equal to the upper half of the hexagon (which is a trapezoid). Why?
(e) It follows that
the area of one of these lunes equals one-third the difference between
the area of the trapezoid and the area of the semicircle that contains
the lune. Why?
Copyright (c) 2000.
Daniel E. Otero