188) Two squares. Find length 'X' | Math Olympiad | Geometry.
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Пікірлер: 8
Awesome!
@relishmath5632
5 ай бұрын
Glad you think so!
10sqrt2 in this way: Rotate the smaller square untill touching the larger square. Call x the side of the larger square and y the side of the smaller one. Applying Pythagorean theorem: X^2+y^2=10^2 For the red line we write: X^2=(x+y)^2+(x-y)^2=2x^2+2y^2=2*(x+y)^2=2*10^2=200 X=10sqrt2
@Okkk517
5 ай бұрын
You can also take x=y to simplify the calculation and to end up with the same result.
@soli9mana-soli4953
5 ай бұрын
@@Okkk517 Great!!
a slight variation of video 111) kzread.info/dash/bejne/hKafx8mxerqtm6Q.html&lc=UgzHIDuy0Il5Kn80V2h4AaABAg
@relishmath5632
5 ай бұрын
Excellent observation
Pytagorean theorem: X² = 10²+10² = 2 . 10² X = 10√2 cm ( Solved √ ) The figure is not defined, can't be drawn. We can choose squares dimensions or rotation angle, in a limit condition. I chose rotate counterclockwise yellow square, making segment '10' vertical. x² = (a+b)² + (a+b)² = 10²+10²