*Tuesday I did some math. Here is the story...*

Tuesday, Oct 7, 8:00 A.M.

Seth Godin blogs every day and Tuesday’s post says that anyone who can read well should be able to do math well. I agree. I never learned to do math well. Lately I have been trying to relearn (or learn for the first time) some math topics. So I admit I’ve had to think hard about Godin’s post’s closing question: "Can an eight-inch square pizza fit on a nine-inch round plate without draping over the edge?”

I’ll share my thought process and answer here without knowing if I am correct. And if I am way off base my math whiz friends like Matt and Zach Miller and Keenan Mann will hopefully correct me.

I assumed an 8 inch square pizza has 4x2 dimensions. Then after thinking for a little while I was fairly confident I had remembered pi times radius squared gives the area of a circle. So I figured a 9 square inch circle has a radius of about 1.7 and therefore a diameter of 3.4 inches.

The 4 corners of the 8 inch square pizza will hang off the 9 inch round plate.

Tuesday, Oct 7, 2:00 P.M.

Keenan’s response:

I agree with the "no", but our math differs slightly. I didn't remember any circle rules/formulas, so my math is based on a quick refresher and therefore could be lacking...

But i assumed the "9 inches around" was describing the circumference. Then I divided that by pi to get the diameter of 2.86. Half that is 1.42, which is the radius. And since pi r squared equals the area of a circle, A=6.44.

So since 8 is the area of the pizza, it wont fit in the circle.

Tuesday, Oct 7, 2:30 P.M.

Although Keenan and I came up with the same “answer,” I didn’t buy that the premise of the question was that the area of the square was given while the circumference (which is like the perimeter) of the circle was given to answer the question. Other than to play a trick, why would apples and oranges be presented, measurement wise? So I typed “Does 10" pizza measure circumference or area?” into google. This link came back. From these educational resources, I conclude that neither Keenan nor I was correct in knowing the premise of the question. Evidently an 8 inch square pizza is a square pizza with four 8 inch sides and a 9 inch round pizza is a round pizza whose diameter is 9 inches.

I then (crudely) drew a square with 8 inch sides and (crudely) drew over it a circle with a 9 inch diameter. I attempted to place the center of the circle in the “center” of the square. The end points of the line segment of the diameter were slightly outside the edges of the square at the “top, bottom, left and right edges" of the circle. But as you can imagine (and hopefully see if I figure out how to post an image of what I drew), because of the shapes of squares and circles, there are many portions of the square pizza that will hang off the edge of the circular plate.

Tuesday, Oct 7, 3:00 P.M.

I searched Google and Twitter for “answers” to “Seth Godin’s math problem” and found this.

This is similar to what I came up with after discovering (I think) the premise of the question.

The thing that gives me pause now is whether, by Mr. Godin’s standards, this is all evidence of my “not being good at math.” He mentioned this question being one that should “make you smile, not one you should have to avoid.” If he meant that there is some clever way to know the answer to this question before pondering, calculating, searching and looking for reassurance, and that if you don’t know that clever way then you are not good at math, then I am not good at math. But if he meant that if you don’t avoid such a question you are either good at math, or that you have what it takes to be good at math, then I agree with his original premise that “all of us are capable of being good at math.”

The problem isn't solved by looking at the total area of either format. To avoid drooping over the side of the plate you need to look at the longest dimension of the square pizza and see if that will fit on the round plate.

ReplyDeleteA nine inch round plate has a diameter of 9 inches. The square pizza must be smaller than 9 inches corner to opposite corner to fit on the round plate without drooping over the edge. To calculate the distance divide the pizza into two right angle triangles by drawing a line from corner to opposite corner. The unknown length is our hypotenuse, the two other sides are 8 inches.

We remember Pythagoras' theorem, the sum of the squares of the two short sides of a right triangle equals the square of the hypotenuse. a(squared) + b(squared) = c(squared). 8 squared is 64, 64+64 is 128, and the square root of 128 is a little over 11.3 (I used a calculator) for that part. So the square pizza, corner to opposite corner, measures 11.3 inches. That's enough for the corners to droop over the side of the round plate.

Uh oh, cheese on the floor.

Thank you @tynan for the explanation and for showing a practical use of Pythagoras' theorem. That type of analysis was not in my toolbox until now!

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