I used a very different algebraic method for the following problem. The math teacher shows two methods in his video, neither of which I used (my solution will be in the comments). I think the teacher's first method and my own method are roughly equivalent in efficiency, but the teacher's second method is short and pretty innovative, being the sort of out-of-the-box thinking I have little aptitude for. Try the problem out, look at my solution, look at the teacher's solutions, and decide which one is best.
Monday, July 03, 2023
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We're given a big rectangle divided into three smaller rectangles by vertical line segments. The area of the skinny, left-hand rectangle is 80; the area of the thicker right-hand rectangle is 176. Our task is to find the area of the center rectangle, marked in yellow.
ReplyDeleteUsing my nomenclature, and looking at how the long horizontal sides are divided into three shorter segments each, let's call the shortest segment, associated with the 80-area rectangle, x. Let's call the medium segment, associated with the 176-area rectangle, y. Let's call the height of all these rectangles z. Finally, let's call the length of the middle rectangle's horizontal side a.
Here's a few things we now know:
32 + y = 40 + x (long side = long side)
x + a = 32
y + a = 40
y - x = 8
xz = 80
yz = 176
Those last two equations are crucial because you can now establish a proportion:
yz/xz = 176/80
The "z"s cancel, so
y/x = 176/80 = 2.2/1
So y = 2.2x! Ha!
Since we know y - x = 8, let's swap in 2.2x for y:
2.2x - x = 8
That becomes 1.2x = 8, so
x = 20/3
The long side of the rectangle is therefore
40 + 20/3, or in decimals, 46.6̄ (i.e., 46⅔).
To solve for z, then, we go
(20/3)z = 80
z = 12
We could solve for y, but at this point, we can calculate the area of the large rectangle, then subtract the areas of the blue and green rectangles to find the area of the yellow rectangle.
The area of the big rectangle is
12(46.6̄) = 560
So the area of the yellow rectangle is
560 - 176 - 80 = 304.
QED.
I'm not totally satisfied with the above explanation because I suspect it includes a few extra, unnecessary steps. But at the beginning, I felt it was a good idea to lay out as many equations as I could even if I didn't end up using them all.