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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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