I have a book of puzzles from 1972 with the pretentious title, "Games for the Superintelligent" by James Fixx. One puzzle had me thinking for a couple of days: I drew it out, thought about different ways of attacking it, and eventually gave up and looked up the answer in the back of the book. See if you can do better than me and figure out the answer on your own!

• Welcome to Puzzling :D – ABcDexter Dec 23 '18 at 20:50
• I feel like this should go on Math.SE. It's just elementary math, rather than a "puzzle". – user46002 Dec 23 '18 at 21:24
• Fun trivia*: Richard Feynman was stumped by a similar problem! (*source: It's a bit of telephone game, as the source is Walter Bender’s webpage which shares a story told by Oliver Selfridge: web.media.mit.edu/~walter/MAS-A12/week11.html) – Presh Dec 24 '18 at 9:19
• This was a lot of fun. I sat down with my wife for ten minutes trying to crack this. We ended up trying to rewrite things in terms of the usual polar coordinates and facepalmed when the answer spilled out. – user1717828 Dec 24 '18 at 18:04
• That was a great puzzle! (Are the other puzzles in the book just as fun? I am going to look for it) – BruceWayne Dec 24 '18 at 23:31

Is it:

8 inches. Because the other diagonal of the same rectangle is also $$8^"$$, which coincidentally, is also the radius of the circle.

• "the rectangle with the 8′′ marked other diagonal is the radius", so basically "the rectangle ... is the radius"? That seem ungrammatical to me. – hkBst Dec 25 '18 at 9:50

Lol! If you're using Pythagorus you're doing it wrong :-)

8" Draw a diagonal line between the other 2 corners of the rectangle. You'll immediately see the answer and then you'll be kicking yourself!

So, A right-angled square rests plumb atop the diameter of a circle. The distance from the square edge to the circle’s edge is 3”, so then the radius of the circle would be the 5” of the square arm, which is at right angles at two arms spread from opposite the hypotenuse in a meeting of two mirrored triangles; added to the 3” of extension to the circle’s edge -resulting in a radius measurement of a combined(square arm+extension to circle edge) 8”. I solved this visually first, but then challenged myself to articulate. I still feel rocky about the explanation and should have drawn it out to clarify as I have been taught instinctual to “show my work.”