Tuesday, August 19, 2008

Back-to-School Geometry - Rectangles, Squares and Deeper Challenges

The connections between geometry and other rich areas of mathematics are boundless. Here is a fairly straightforward set of problems that can be explored as far as your eye and mind can see. On the surface, we have three rectangles each of which has a half-diagonal of length 6. Students can be asked to find the area of each without using any trigonometry. On a deeper level, one can ask students to explain or prove why the square has the greatest area for a given
diagonal length. This is straightforward using the well-known trig formula for area of a triangle, K = (1/2)absin C, however the challenge here is to use non-trig methods (although the student can use special right triangles) to compute the areas and demonstrate the maximum. The maximum piece is more sophisticated and the idea of bringing this in before precalculus and calculus has many benefits.

The instructor might begin by asking students to draw any rectangle with a diagonal of 12. How many such rectangles could there be? Which one would appear to have the greatest area? Ok, now let's explore a few special cases.

This problem allows the creative student to devise a visual way of explaining the maximum. It also allows the instructor to bring in the Arithmetic Mean-Geometric Mean Inequality for enrichment. So many methods and approaches are possible...


Totally_clueless said...

One variant would be to prove that the perimeter is also maximized when the rectangle is a square.

It appears that the above problem can be reduced to the area maximization prblem.


Dave Marain said...

Nice, tc!
For a given diagonal length, the square maximizes both the area and perimeter, an interesting result to say the least. In some ways, I like the perimeter. Students might imagine that we could increase the perimeter by making a long thin rectangle but the diagonal constrains this. The perimeter result can also be demonstrated using the AM-GM ineq, calculus or by other methods.