You guys. Crazy story. I was in a coffee shop messing around with Geometer’s Sketchpad, just slapping together some perpendicular line segments because that’s how I roll, and this guy who could see my screen thought I was an artist or something. It was kinda weird, but he wanted to buy my “art.” Naturally, I went along with it. So now I have to frame this stupid thing and deliver it to him. Thing is, he was really particular about how it should be framed. He wants both points A and J to be on one of the rectangular frame’s diagonals, and a distance of exactly 4 cm along those diagonals from A and J to the inner corners of the frame. He wants the frame to be 3 cm thick, too. Anyway, he’s busy clearing space for this thing in his house and he’s emailing me asking for dimensions and basically driving me crazy. Can you help me out? What will be the outer dimensions of the framed piece of art? I’ve included measurements of all the lengths of the segments, but don’t count those as part of my image—I’m going to delete them before I print this out.
Post your answers—rounded to 2 decimal places—in the comments. The prize for the first correct answer: I will name my next pet goldfish after you. Scout’s honor.
UPDATE: Here’s where I sheepishly admit that this is not one of my best challenges. Sorry. But Rob finally did get it, and even though I didn’t plan to when I posted this, I’m gonna send him a Math Guide for his troubles.
Solution below the cut.
The first thing you’re gonna want to do is construct a right triangle out of this bad boy, which means you’re going to need to find the total vertical length and the total horizontal length. What a pain!
Vertical length (AK) = AB + CD – EF – GH + IJ = 12.87
Horizontal length (JK) = BC + DE + FG + HI = 12.88
So that’s super annoying and why this is not one of my best challenges—Geometer’s Sketchpad is an awesome program that I love and it makes making diagrams really easy and fun, but this was drawn up to be a 45º-45º-90º triangle and because of rounding now it’s not quite so pretty. We’ll be OK, but still. Annoying.
Anyway, now we can find AJ, the hypotenuse.
12.872 + 12.882 = AJ2
331.5313 = AJ2
18.2080… = AJ
OK, good enough. We’re told in the problem that the inner corners of the frame are to be 4 cm from A and J. How the heck do we do that?
Easy, friend. Similar triangles. Picture it with me. You’re going to be creating another, larger, right triangle when you extend that hypotenuse by a total of 8cm. So with a few quick ratios, we can determine the inner dimensions of the frame!
The hypotenuse of our large triangle is going to be 18.2080… + 8 = 26.2080…
Do some ratios to figure out the height and width:
But we’re almost done. Now all we need to do is account for the width of the frame, which is supposed to be 3 cm. Because the frame’s thickness will add to top and bottom, left and right, we add 6 cm to each side to arrive (finally) at our outer dimensions: 24.52 cm by 24.54 cm.