652. Three square units extreme

Credit: The Moscow puzzles (1972) Boris A. KordemskyThe 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

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We must now proof that the area of the parallelogram is 3 square units. To calculate the height:sinš¯›³ = height/1height = sin(36.869898)height = 0.6 unitThe area of the parallelogram is:= base X height= 5 X 0.6= 3 square units

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