According to Martin Gardner, a popular American mathematics and science writer, this puzzle was invented by a New York City amateur magician Paul Curry in 1953. Compare these two triangles below.
Both triangles, the first (Triangle 1) and second (Triangle 2), depict two arrangements of shapes, each of which apparently forms a 13 unit × 5 unit right-angled triangle, but Triangle 2 has one missing square in it.
Both Triangle 1 and Triangle 2 are made up of the same four components, namely:
1) Red right-angled triangle with a measurement of 8 unit x 3 unit.
2) Blue right-angled triangle with of measurement of 5 unit x 2 unit.
3) Green L-shaped figure consisting of 8 square unit.
4) Yellow L-shaped figure consisting of 7 square unit.
However, notice that Triangle 2 has one missing square.
How can this be?
The key to the puzzle is the fact that neither Triangle 1 nor Triangle 2 has the same area as the combined area of its components.
Let’s measure the areas of the four components:
1) Red right-angled triangle. Area = 0.5 x 8 x 3 = 12 square unit
2) Blue right-angled triangle. Area = 0.5 x 5 x 2 = 5 square unit
3) Green L-shaped figure. Area = 8 square unit
4) Yellow L-shaped figure. Area = 7 square unit
Combined area of the four components = 12 + 5 + 8 + 7 = 32 square unit
BUT
Calculated area for Triangle 1 (or Triangle 2 if you ignore the missing square) = 0.5 x 13 x 5 = 32.5 square unit, or so it seems.
So, the combined area of the four components does not tally with the calculated area of Triangle 1 or Triangle 2.
So what does this mean?
The red triangle has a ratio of 8:3 while the blue triangle has a ratio of 5:2. This means these two hypotenuse lines do not have the same gradient. So the apparent combined hypotenuse in both Triangle 1 and Triangle 2 are actually bent. In other words, the hypotenuse in the red triangle is not parallel (in the same straight line) as the hypotenuse in the blue triangle for both Triangle 1 and Triangle 2.
Note the grid point where the red and blue hypotenuses meet in Triangle 1, and compare it to the same point in Triangle 2; the edge is slightly over or under the mark. Overlaying the hypotenuses from Triangle 1 and Triangle 2 results in a very thin parallelogram with the area of exactly one square, the same area “missing” from Triangle 2.
Got it?
Source: wikipedia.org, marktaw.com
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