A cube whose length of one side is 6 cm is put on the flat floor as shown in the figure.
This cube is rolled to the direction of the arrow with putting the side of FG on the floor so as not to be slipped until the face BFGC reaches on the floor.
(1) Find the area of the part where the line segment AF moved.
(2) Find the volume of the part where this cube moved.
Answer
(1) 56.52 cm2
(2) 555.12 cm3
Solution
(1) As shown in the figure below, shadow portion is the area AF moved.
AF × AF / 2 = 6 × 6 = 36.
AF × AF = 72.
The area AF moved = AF × AF × 3.14 × 90/360 = 72× 3.14 × 90/360 = 56.52 cm2.
(2) The area of the bottom of the part the cube moved is 56.52 + 6 × 6 = 92.52 cm2.
Therefore the volume is 92.52 × 6 = 555.12 cm3.
