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Credit: The Moscow puzzles (1972) Boris A. Kordemsky
The length of a matchstick is 1 unit, so in the layout above 12 matchsticks were used to create an area "W" of 3 square units. (see puzzle 121 and also how area W was calculated below)
There is another way to create 3 square units, but you may NOT MOVE the matchsticks which from line AC. Can you find it?
Take note of the calculations below as you might need some of the information in your solution.
The area W was calculated as:
= Total area ABC - area X - area Y - area Z
= 1/2(3 X 4) - 1 - 1 - 1
= 3 square units
Angle š¯›³ was calculated as:
= arcsine(AB/AC)
= arcsine(3/5)
= 36.869898 degrees
We must now proof that the area of the parallelogram is 3 square units.
To calculate the height:
sinš¯›³ = height/1
height = sin(36.869898)
height = 0.6 unit
The area of the parallelogram is:
= base X height
= 5 X 0.6
= 3 square units
To calculate the height:
sinš¯›³ = height/1
height = sin(36.869898)
height = 0.6 unit
The area of the parallelogram is:
= base X height
= 5 X 0.6
= 3 square units
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