I found this problem online yesterday (briefly) via a post on the google+ k-12 education. In the figure below

- $\bigtriangleup$ BCX has an area of 5
- $\bigtriangleup$ AXY has an area of 3
- $\bigtriangleup$ CDY has an area of 4

What is the area of $\bigtriangleup$ CXY in the middle? So I have a reasonable solution for the problem but I'm still trying to see if I can find a way to eliminate the quadratic equation from the process.

### My solution

First add in some perpendicular lines and divide the big rectangle into 4 quadrants.

Conveniently we know most of the quadrants areas from the triangles.

- $\Box$ AXOY has an area of 6 since its two $\bigtriangleup$ AXY.
- $\Box$ BCXZ has an area of 10 since its two $\bigtriangleup$ BCX.
- $\Box$ CDYW has an area of 8 since its two $\bigtriangleup$ CDY.

Notice the last two boxes overlap in the upper right corner and I split that out into x. Now we have 2 sets of proportional rectangles going either horizontal or vertical. I'll just choose the vertical ones but you can go either way.

$\frac{\Box BWOX}{\Box AXOY} = \frac{\Box WCZO}{\Box OZDY}$

so

$\frac{10-x}{6} = \frac{x}{8-x}$ which once simplified produces the following quadratic:

$x^2 - 24x + 80 = 0$ or $(x - 4)(x - 20) = 0$ We choose the first solution where x = 4 since the second produces negative areas in the other boxes. That means the entire rectangle has a size of

6 + 6 + 4 + 4 = 20 and

$\bigtriangleup$ CXY = Area of the big rectangle minus the 3 triangles or 20 - ( 3 + 4 + 5) = 8.

#### Improvements?

The presence of the quadratic means I couldn't give this one to the kids but it seems like there should be a refinement that lets me remove it. I thought about it a bit yesterday and couldn't come up with any simplifications. Jump in if you have any ideas.

You can certainly do this by guess and check.

ReplyDeleteTake advantage of the fact the aspect ratio is under-specified, so you can just pick them. Make the rectangle 5 units wide. That tells you the height of the 5 triangle is 2. Then you can pick the 3 triangles is 2 units high and 3 wide. The 4 triangle works out at 4 units high and 2 wide. This is a guess & check solution - you are essentially guessing the answer on the quadratic.