Problem R-1 - solution

SOLUTION TO PROBLEM R-1

First some notations. Let ABCD be the trapezoid such that AB and CD are parallel, AB = a, CD = b, BC = x, AD = y. Let E be the point on CD such that ABCE is a parallelogram. So AE = x and DE = b-a. The area of the triangle ADE is given by

$\begin{displaymath}{1\over 2} xy\sin {\pi \over 4} = {1\over 2}(b-a)h \end{displaymath}$

where h is the height of the trapezoid. Using the cosine theorem we get

$\begin{displaymath}(b-a)^2 = x^2 + y^2 - 2xy\cos {\pi \over 4}. \end{displaymath}$

Thus

$\begin{displaymath}(b-a)^2 = x^2 + y^2 - 2(b-a)h. \leqno(*) \end{displaymath}$

Applying the Pythagorean theorem in the four right triangles AOB, BOC, COD, DOA (where O is the intersection of the diagonals), we obtain that

x² = OB² + OC²

y² = OD² + OA²

a² = OA² + OB²

b² = OC² + OD²

Therefore

x² + y² = a² + b²

and by replacing this last equality in (*) and performing the computation we get

$\begin{displaymath}h = \frac{ab}{b-a} \end{displaymath}$

Questions and/or comments should be directed to Judy Downey or Griff Elder

Back to the Problem of the Week Page

Last modified: Fri Jan 18 19:52:59 CST 2002