To Accumulate a Rate — Integrate!

Teaching High School Mathematics

Area of a Frame?

A guy that teaches math a couple doors down from me sent me a picture tonight of his son’s math problem from a test or quiz. Here it is:

perimeter1

Then he called me and we had a good discussion about the problem. In my mind I wanted to find the area of the shaded part that makes up the wooden part of the frame. He wanted to find the area of the whole thing minus the 3 rectangles where the pictures go.

Note that this is in a 6th grade class on pre-Algebra.

Let’s call the width of the shaded frame w.  The width, w= (l – (4x+15) )/2. Then 10w times 2 = 20w for the two vertical parts of the frame and w times l times 2 = 2wl for the two horizontal parts of the frame. But we included all the corners twice and so would need to subtract them. -4w^2. My expression would be 20w +2wl -4w^2 using the above expression for w.

My friends option is a problem. It would be 10l – 3 times 5 times ? There is no way to find the height of the pictures.

I figured that there was no way they wanted students to do the problem my way, and it was impossible to solve it the other way… What did they want?

I typed the question into google and the book name and I got a similar problem… Here it is:

perimeter2

The answer to this problem is obviously A = 6(a+2). After looking at this one, I’m thinking that when they say frame, they mean the area of the whole thing.

This makes the area of the frame above = 10l. Done. And I don’t like it. I’m fine with the extra info you don’t need. I’m not fine with them asking for the area of a frame and not explaining what they mean by the area of a frame and shading in a part that couldn’t possibly be what they want you to find the area of.

Update: The teacher shared the answer the next day and wait for it…. 10(4x+15) Ahhhhh!!!

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This entry was posted on November 2, 2016 by in Uncategorized.
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