# Side of the octagon obtained by intersecting a square and its image by a $\pi/4$ rotation

2 squares of side 2x overlap to form a regular octagon. How long is each side of the octagon? Image of question: http://www.ucl.ac.uk/language-centre/placement-tests/UPC/Maths/images/question14.jpg The hint they gave is: Try again. If the middle piece of each side of the square is y, we get a right-angled triangle all of whose sides can be expressed in terms of x and y. Then apply Pythagoras's Theorem.

may you please explain, I do not get it at all. I saw the question posted already, howevr I cannot comment as i still do not have 50 reputations, also the answer they gave it was ununderstandable. may u plz provide a drawing. thank u

• Here it is: math.stackexchange.com/questions/2072973/… – Rohan Jan 27 '17 at 13:35
• i know howeevr, there is no drawing and their answer is unclear. and i cannot comment because i dont have enough reoutations – exchangehelpforuni Jan 27 '17 at 13:36
• @rohan look above – exchangehelpforuni Jan 27 '17 at 13:37
• One answer has been deleted which gives more explanation that the existing one. Can I post that?? – Rohan Jan 27 '17 at 13:37
• i dont know its upto u if it helps @rohan – exchangehelpforuni Jan 27 '17 at 13:39

Each side of the octagon is equal to $2x-\sqrt{2}x$. • wait what did u jsut do here???? why is ab=sqrt2x and everythign else????? how did u get these numers u should explain – exchangehelpforuni Jan 27 '17 at 15:44
• @exchangehelpforuni, isn't $ABC$ a right angled triangle? I just used pythagorean theorem. Therefore $AB=\sqrt{2}x$ – Seyed Jan 27 '17 at 15:53
• how did u use pytahgorean theorm if u only know what one side is? – exchangehelpforuni Jan 27 '17 at 15:57
• nevermind i got it i was being stupid – exchangehelpforuni Jan 27 '17 at 15:58
• plz calrify how u got ad... i know u did ab-x, but what is the x – exchangehelpforuni Jan 27 '17 at 16:00

With the OP's permission:

Let each side of either square be $2x$. This is divided into two pieces of length $y$ outside the second square and a piece of length $z$ inside the second square. Thus:

$2x=2y+z$

At each corner of any square is a right triangle with legs $y, y$ and hypotenuse $z$. Now apply the Pythagorean Theorem:

$y=(z\sqrt{2})/2$

Now eliminate $y$ and get the relationship between $x$ and $z$. The answer follows directly.

Hope it helps.

• @rohan thank u v much. however i do not get why z is the hypotenuse... – exchangehelpforuni Jan 27 '17 at 15:50