MO SK/CZ Z9–I–3

John had the ball that rolled into the pool and it swam in the water. Its highest point was 2 cm above the surface. Diameter of circle that marked the water level on the surface of the ball was 8 cm. Determine the diameter of John ball.

Result

D =  10 cm

Solution:

$r^2 = (8/2)^2+(r-2)^2 \ \\ r^2 = 16+r^2-4r+4 \ \\ 0 = 20-4r \ \\ r = 20/4 = 5 \ cm \ \\ D = 2 \cdot \ r = 2 \cdot \ 5 = 10 = 10 \ \text{ cm }$

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Math student

I need more explanation as to where these equations are coming from.

2 years ago  Like

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