a sector as shown in the figure is rolled up to form a right circular cone .'
the figure"
http://i17.photobucket.com/albums/b93/mindrew715/Picturekj56.jpg
find the base radius and the volume of the cone ...
2007-02-01 9:43 pm
回答 (2)
2007-02-03 8:07 pm
✔ 最佳答案
圖片參考：http://i17.photobucket.com/albums/b93/mindrew715/Picturekj56.jpg
The sector as shown is rolled up to form a right circular cone.
So, the length of the sector will be equal to the circumference of the base of the circular cone.
The length of the sector = 2π(5) * (216/360) = 6π
Let r be the radius of the base of the circular cone.
2πr = 6π
r = 3
The radius of the base of the circular one is 3 cm.
Next, let h be the height of the circular cone.
By Pythagoras' Theorem, h² + 3² = 5²
h² = 25 - 9
h = 4 (-4 is rejected)
Therefore, the volume of the cone is:
1/3 πr²h = 1/3 π 3² * 4 = 12π cm³
(or 37.70 cm³ correct to 2 d.p.)
2007-02-05 6:23 am
The sector as shown is rolled up to form a right circular cone.
So, the length of the sector will be equal to the circumference of the base of the circular cone.
The length of the sector = 2π(5) * (216/360) = 6π
Let r be the radius of the base of the circular cone.
2πr = 6π
r = 3
The radius of the base of the circular one is 3 cm.
Next, let h be the height of the circular cone.
By Pythagoras' Theorem, h² + 3² = 5²
h² = 25 - 9
h = 4 (-4 is rejected)
Therefore, the volume of the cone is:
1/3 πr²h = 1/3 π 3² * 4 = 12π cm³
So, the length of the sector will be equal to the circumference of the base of the circular cone.
The length of the sector = 2π(5) * (216/360) = 6π
Let r be the radius of the base of the circular cone.
2πr = 6π
r = 3
The radius of the base of the circular one is 3 cm.
Next, let h be the height of the circular cone.
By Pythagoras' Theorem, h² + 3² = 5²
h² = 25 - 9
h = 4 (-4 is rejected)
Therefore, the volume of the cone is:
1/3 πr²h = 1/3 π 3² * 4 = 12π cm³
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