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A square and its lines...

Math brain teasers require computations to solve.


Puzzle ID:#193
Fun:** (2.01)
Difficulty:*** (2.73)
Submitted By:trojan5x***




Imagine a square that has each side consisting, half of length `a` and half length `b`. Inside the square there is a diamond, with each corner touching the midpoint of each side of the square. Each side of the diamond has length `c`. If one were to set the area of the square equal to the area of all the shapes inside the square, what would remain?

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The Pythagorean Theorem.

The area of a square is `length squared`. First find the area of the entire square. The length is (a+b). The area of the square is (a+b)^2. Inside the box there is a diamond with length `c`. The diamond is simply a square tilted 45 degrees. The area of the diamond is c^2. There are also 4 triangles with area 1/2ab. Set them equal:
(a+b)^2=c^2 + 4(1/2ab) Expand...
a^2 + 2ab + b^2= c^2 + 2ab
Subtract `2ab` from both sides and you are left with the Pythagorean Theorem.
a^2 + b^2 = c^2


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Jul 02, 2004

Awesome, proving the pythagorean theorem was an extra credit assignment in my geo class this year, and i couldn't figure out how to do it without using the way we did it in class, but this could have worked! cool!
Jul 13, 2004

This may prove the Pythagorean Theorem (for which there are dozens of proofs) but more importantly, it proves that: ONE PICTURE IS WORTH 1000 WORDS.
May 25, 2005

Jan 03, 2006

One picture worth a thousand words woyuld be useful right now beacause I didn't follow the statements or the question.
Jan 10, 2006

I honestly didn't understand what the question was asking for, either. I understood the description of the setup perfectly, but didn't know what you meant by "if you set the area of the square equal to the area of the things inside..." The area of the square HAD TO BE the same as the total area of the shapes inside. After all, the shapes inside are defined by and contained in the square.

Interesting derivation of the Pythagorean Theorem, though, and I must give kudos for that!

Jan 20, 2006

great teaser but...huh?
Jan 21, 2006

Sorry, this "puzzle" doesn't do it for me. As to the Pythagorean theorem, the demonstration focuses on the single case where A=B and is not a general proof. Instead, one can divide the sides of a square into segments of lenghth A and B; the ratio A:B can be anything, not just 1:1. (The B segment of every side must be clockwise from the A segment, or vice versa.) Create four triangles (Each will have area of A*B/2.) and the smaller square (with sides of lenghth C) by joining each dividing point between A and B segments to the dividing points on adjacent sides. Either picture these triangles folded (or actually fold them) inward so they cover part of "square C." You will find you create a new square with sides the lenghth of the difference between A and B. The area of the C square is equal to the new square plus the area of the four triangles. C^2 = (A-B)^2 + 4(A*B/2). C^2= A^2 - 2A*B + B^2 + 2A*B. C^2 = A^2 + B^2.
Apr 03, 2006

I agree with stil
Apr 05, 2006

i never thought i'd say this about math but WHOA! THATS SO AWESOME!!!!!!!!!!!!!!!
Apr 13, 2006

it was a square, huh? so a = b.. is that correct? i don't really understand the question.. but .. nice.. althou i've seen this one (not in text.. but in picture).. lov it
Apr 25, 2006

hey! we studied that!!
May 17, 2006

I didn't understand what the question was asking.
Jun 14, 2006

This is actually the converse of the Pythagorean Theorem. a² + b² = c² is the converse. c² = b² + a² is the Pythagorean Theorem. But hey, in Algebra I we came up with a=b iff b=a.
Jul 16, 2006

we had to do that in math last year.

and that is the Pythagorean Theorem, not the converse
Aug 23, 2006

I'd say nice teaser, but your wording really didn't make sense... sorry.
Oct 13, 2006

hard to understand what u were talkin bout.
Dec 19, 2006

The area of the triangle would be ((a+b)^2)/8.

base of triangle = (a+b)/2
height of triangle = (a+b)/2
because the diamond intersects the square at the midpoint of each side.
Dec 19, 2006

just FYI-- A does not have to equal B in this situation, except for the fact that the setup calls for a "diamond" in the center, which suggests congruency along at least one center line. in this case, since it's a diamond within a square, at the midpoint of each side, it is indeed also a square. BUT, for any square with 4 sides equalling A+B (and, going around the square, never B+A) and building a shape in the middle connecting the point between A & B on each side, the resulting polygon would always be a RHOMBUS-- a quadrilateral with equal sides. in this case, the sides of the polygon in the center would always = c, the area would equal c^2, and it would be a proof for the pythagorean theorem even if a b. Causes a problem tho if the sides were ever flipped so that two A segments are in the same corner (and two B segments would thus have to occupy at least one other corner.) this twist would mean that a) the quadrilateral in the center would not be equilateral, and the area would be more difficult to compute. BUT, it would still have the same area.
Feb 07, 2007

poorly, poorly, POORLY worded, i am sorry to say.
Feb 13, 2007

I got this right and didn't even know it! I agree with a lot of people, that this was poorly worded. But I drew out the square and the diamond, and wrote a^2 + b^2 = c^2?? under it. I then got confused by the wording of the question and gave up, but when I looked at the answer, hey! I was right!
Feb 20, 2007

C^2 also equals 2b^2 or 2a^2...
Sep 08, 2008

it seems clear to me the reason this teaser is supposedly poorly worded (may be the case but easy enough to understand) is simply because the writer knew the answer before the question, maybe saw a proof for pythagoras' theorem and wanted to post a teaser on here demonstrating it.

This is definitely my favourite proof for pythag, logical and simple and immensely satisfying, i first came across it in "Fermat's last theorem (simon singh),"and the fact that a=b doesn't matter because they're always referred to as a or b, that fact that they are the same doesn't alter the algebra involved, you don't need that assumption so this is a perfectly fine example of proof for pythagoras' theorem.
Jan 16, 2009

Horrendously worded.

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