Problem 8 (Trapezoid Problem)

Consider the following trapezoid with parallel sides of 20 inches and 17 inches, and an altitude of 2 inches, where the angle at vertex A is 45 degrees.

Can you cut this trapezoid into smaller congruent figures ?

Solution

Since the angle DAH₂ = 45 degrees and the altitude of the trapezoid is 2 inches, we have that DH₂ = AH₂ = 2 inches. Hence, we can conclude that H₁B = 1 inch, which makes the angle CBH₁ = 60 degrees. Hence, there appears no hope in subdividing the trapezoid into congruent triangles since the triangle ADH₂ has equal legs whereas the triangle ABH1 is a 30-60 degree right triangle, and it is not difficult to show neither triangle can be broken into smaller congruent regions. Hence, we attempt to subdivide the trapezoid into smaller congruent trapezoids as illustrated in the following figure.

One thing is now imediately clear. If we can subdivide the trapezoid into smaller congruent trapezoids, the number of these trapezoids must be an odd integer 3, 5, 7, .... . Letting n be the number of trapezoids, we can add the length that these trapezoids lie along the line AD for different numbers n, getting

number of trapezoids	total length of AD = 20
... 3 ...	2(x + 3) + x = 20
... 5 ...	3(x + 3) + 2x = 20
... 7 ...	4(x + 3) + 3x = 20
... 9 ...	5(x + 3) + 4x = 20
...	2 (x + 3) + x = 20
... n ...	((n + 1)/2) (x + 3) + (n - 1)/2) x = 20

Solving this last equation for x in terms of the number of trapezoids n, we get

x = (37 - 3n)/2n

for n = 3, 5, 7 ... . For these values of n the only positive values of x are

number of trapezoids n	3	5	7	9	11
size x of the trapezoid	14/3	11/3	8/7	5/9	2/11

Hence, we can subdivide the trapezoid into smaller congruent trapezoids with 3, 5, 7, 9, or 11 trapezoids. The subdivision into 3 trapezoids with size x = 14/3 is shown in the following diagram.

Last modified on Tuesday, January 12, 1999