Danny Low's answer to tiffany's Secondary 4 E Maths Singapore question.
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Hope this helps
Date Posted:
2 years ago
the ball should contact the wall, floor AND cylinder, so ED is not diameter of ball.
let radius of ball be r, then
(sqrt[2] x r) + r = ED
r (sqrt[2] + 1) = 2 sqrt[2] -2
r= 6-4sqrt[2]
= 0.34314
volume of ball
= (4/3) pi (0.34314)^3
= 0.16924 ...
= 0.17 m3 (to 2 d.p.)
have tried to draw a scaled diagram of a cylinder against floor and wall, and a ball squeezed into the corner. from the diagram, the radius of the ball corresponds quite closely to the calculated radius of 0.343145 above.
the exact value of r above is obtained by simplifying the surds. for students who do not do A maths, can still find r by converting the surds to the corresponding decimals.
let radius of ball be r, then
(sqrt[2] x r) + r = ED
r (sqrt[2] + 1) = 2 sqrt[2] -2
r= 6-4sqrt[2]
= 0.34314
volume of ball
= (4/3) pi (0.34314)^3
= 0.16924 ...
= 0.17 m3 (to 2 d.p.)
have tried to draw a scaled diagram of a cylinder against floor and wall, and a ball squeezed into the corner. from the diagram, the radius of the ball corresponds quite closely to the calculated radius of 0.343145 above.
the exact value of r above is obtained by simplifying the surds. for students who do not do A maths, can still find r by converting the surds to the corresponding decimals.
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