Geometry Puzzle: If the Area of the Blue Square is 25 Find the Area of the Red Square
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- Опубліковано 6 вер 2024
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General solution:
area of large square = a^2
area of smaller square = x^2
then
(a/2 + x)^2 + (x/2)^2 = (a√2/2)^2 (expand the squares)
(a^2) / 4 + a x + x^2 + (x^2) / 4 = 2 * (a^2) / 4 (multiply by 4)
a^2 + 4 a x + 5 x^2 = 2 a^2 (subtract 2 a^2)
5 x^2 + 4 a x - a ^2 = 0 (abc quadratic equation)
x(1,2) = (-4a +/- √((4a)^2 - 4 * 5 * (-a^2)))/ (2 * 5) = (-4a +/- √(16a^2 + 20 a^2))/ 10 = (-4a +/- √(36a^2))/10
x = (-4a + 6a)/10 or x = (-4a - 6a)/10
x = 2a/10 or x = -10a/10
x= a/5 or x = -a
The second answer (x = -a) would give us the blue starting square, so that kind of makes sense.
The first answer shows that there will always be a 5 times reduction in side of the square, and consequently a 25 times reduction in area of the smaller square.
Cna we say that the area of left and right white part in lower of side of blue one is equal??
If yes then answer might be different
Small squares are equal. Side equal to half of radius dia - 5. Any flaw in my argument?
Red area =1^2=1 square units
Chord theorem
Let side of the small square = x
x(5+x)=(2.5-x/2)(2.5+x/2)
x(5+x)=1/2(5-x)1/2(5+x)
4x(5+x)=(5-x)(5+x)
20x+4x^2=25-x^2
5x^2 + 20x - 25 =0
x=1
A=x^2=1 ***
Call s the side of the small square, then Thales and Pythagoras tell us that:
(5 + 2s)^2 + s^2 = (5*sqrt(2))^2
5s^2 + 20s -25 =0
5(s+5)(s-1) = 0
Hence s = 1
Area is 1.
А=1.