Here’s a great problem relating parallelograms and area -

A parallelogram has 10-inch and 18-inch sides and an area of 144 square inches.
a. How far apart are the 18-inch sides?
b. How far apart are the 10-inch sides?
c. What are the angles of the parallelogram?
d. How long are the diagonals?

OK, what’s the pre-existing knowledge or discussion that has happened before the problem is presented to the students? So far, the students have discussed the fact that the area of a parallelogram is simply the base times the height (or altitude, perpendicular distance between the parallel sides), but have not done a huge number of repetitive area problems, I’d say not even two or three of them. They’ve done a dissection discussion problem a few pages back about how to cut off a right triangle and move it over and the parallelogram basically becomes a rectangle, and that’s why it’s the same area formula.

So, in my class, most students got part a, pretty quickly, just divinding 144 by 18 and getting 8. The other type of previous problem the students have done is being given the area of a triangle and all the sides, using each side in turn to find each of the three altitudes, so they area used to turning the triangle around to find each of the heights. So now we are seeing if they can transfer that method to a parallelogram in this problem. This took getting all students to the board in my class.

Let me back up for a second. We also had covered problems that had discussed both the tangent and sine functions for right triangle trigonometery (which are introduced before cosine in our curriuculum). Hence, the students were desperately trying to find the height to the side of 10 using a right triangle, which could not work, of course, without being able to break up the side of 18. Some of them guessed lucky making a 6-8-10 triangle, which did get them the right answer, but they could not justify it – no beans in my classroom.

So, we finally had to get them to draw the parallelogram with the 10 as the base, then students finally realized they could use the area once again. This was the a-ha moment and it spread like wildfire. Thank God. Once we got through this point, they were able to find the angles with the trig no problem.

However, finding the diagonals took some manipulating of right triangles with the heights. The shorter one was easier with the right triangle within the parallelogram. This created a right triangle with legs of 8 and 12 and the hypotenuse was the diagonal. Putting a right triangle outisde the parallelogram made a bigger right triangle with legs of 18 and 8 and that hypotenuse was the longer diagonal. Since I expected so many students to not have totally completed this problem, but only have attempted it, I made time for this to be done at the board in pairs during classtime. It was time well spent, and many students wrote this problem up in their journals – very well, I might add.

If you’d like to try this problem in class it was p. 47 #2.