Ok, geometry was a while ago. My 6th grader has this problem. The large parallelogram is made up of four identical parallelograms. The dimensions are given, as is the height of the parallelogram, in red. She is supposed to find the area of X, which I took to mean the area of the big, outside parallelogram minus 4X the area of the smaller parallelograms. I feel like I'm missing one needed number. I don't know the height of parallelogram X. I don't know any of the corner angles. How do I solve this? I can do it with trig or pythagorean's theorem, neither of which they've learned yet. My 18 year old also solved it using similar triangles. But I don't think 6th grade class does this.

Is there a really simple way to do this and I'm overthinking it?

This is my very own MS Paint masterpiece. It's not to scale.

Without trig I am going with 4 for the inner part.

Large parallelogram is 8x8 subtratct 3 from all four sides makes X a 2"x2" parallelogram.
Since the height given for the side measuring 3" is 2.4", the height of the 2" is (2.4"/3")x2" = 1.6"
A=bh = 2"x1.6" = 3.2 sq. in.
ETA: the angles are not important. The solution is in the ratio of the length of the sides. In this case,  2/3.

An example of how not getting to use something on a regular basis leads towards losing common sense approaches.

Thanks.

Am I missing something here? Isn't this just basically a subtract one area from another area problem?

The large parallelogram has dimensions of 8 x 8.

The area enclosed by that parallelogram is 64 (you can get that by lopping off an imaginary triangle at the left and adding it back on the right, making it into a rectangle with dimensions of 8 x 8.

So the area of the large parallelogram is 64.

Then, we are told that each of the four smaller parallelograms are equal (assuming that they are leaving out the parallelogram marked by X).

So, each of those four smaller parallelograms can be transformed into a rectangle by the same method - that rectangle would have dimensions of 5 x 2.4, for an area of 12. Since there are four of them, the total area of those smaller parallelograms is 4 x 12, or 48.

So to get the area of X, subtract 48 from 64, giving X an area of 16.

This is how I would do it.  Something seems a bit wonky with that approach though as it would be 8 on the top and bottom after moving the triangle, but because it was a parallelogram the height will be shy of 8 by a little.

ETA CAD confirmed my suspicions.  I'm out...

@targetworks
The trick is the height of the parallelogram is 2.4 not 3.
Then you figure out the sides of the rhombus in the middle.
Third you multiply the height by the ratio of the length of the parallelogram sides.

Many thanks for the thoughts. I did think about the ratio approach. I asked my 6th grader if they'd solved some other ones in class that way and she didn't think so.

I'll have her use that method and then ask the teacher about it on Tuesday.

It reminded me of sitting in college freshman physics and doing some trajectories or kinematics and thinking, "This is much easier with a little calculus."

Show your 6th grader Presh Talwalker's youtube channel Mind Your Decisions. After doing all those problems, she will know all the tricks.

I will investigate this. Thanks.

The large parallelogram is 8 wide, and 8 * (2.4/3) tall, or 8 x 6.4.  The total area of the parallelogram is therefore 51.2.

Each individual parallelogram is 5 * 2.4, or 12.  All 4 are therefore 48, and the parallelogram in the center is 51.2 - 48, or 3.2.

Mike

the inner square is 2x2

I think that I erred in using the height of 2.4 in the calculation of the area of the smaller parallelograms.

The area of a 3 by 5 parallelogram is 3 x 5 or 15 (the same area calculation as would be used for a 3 x 5 rectangle).

So four such parallelograms have a total area of 60, the 8 x 8 parallelogram has an area of 64, subtracting one from the other leaves X with an area of 4.

That calculation only works for squares and rectangles.  You were correct in using the 2.4 as the height of the smaller parallelogram.  The trick is that the larger parallelogram is also shorter by the same fraction (2.4/3), so the height of that is 8 * .8, or 6.4.  You can use that to figure out the area of the larger parallelogram (technically, it is a rhombus, which is a parallelogram with all 4 sides being the same length), and then subtract the area of the 4 smaller parallelograms.

Ah, yes, of course - remind me to not attempt geometry before having my morning cup of coffee... (whether it's in the morning or afternoon or evening)

A parallelogram is a square with deflection