A, An equilateral triangle of side length 6sqrt3 is rotated about an altitude to form a cone. What is the volume of the cone?
B, A 288 degree circular sector with radius 15 is rolled to form a cone. Find the volume of the cone.
C, What is the radius, in inches, of a right circular cylinder if the lateral surface area is 24 pi square inches and the volume is 24 pi cubic inches?
D, At the MP Donut Hole Factory, Niraek, Theo, and Akshaj are coating spherical donut holes in powdered sugar. Niraek's donut holes have radius 6 mm, Theo's donut holes have radius 8 mm, and Akshaj's donut holes have radius 10 mm. All three workers coat the surface of the donut holes at the same rate and start at the same time. Assuming that the powdered sugar coating has negligible thickness and is distributed equally on all donut holes, how many donut holes will Niraek have covered by the first time all three workers finish their current donut hole at the same time?
(D has been answered before with the solution 144 donut holes but this is not the correct answer unfortunately)
+128475 A, An equilateral triangle of side length 6sqrt3 is rotated about an altitude to form a cone. What is the volume of the cone?
The radius of the cone is 6√3 / 2 = 3√3
The height of the cone is (√3) * 3√3 = 9
So....the volume of the cone is
(1/3) pi * radius ^2 * height =
(1/3) pi * ( 27) * 9 =
81 pi units^3
+128475 B, A 288 degree circular sector with radius 15 is rolled to form a cone. Find the volume of the cone.
The sector radius will from the slant height of the cone.
The arc length of the sector will form the base circumference of the cone...its length is :
2 pi ( 15) (288/360) = 24pi units
So....the radius of the cone is figured as
24 pi = 2 pi (r)
12 = r
And the height of the cone is
√ [15^2 -12^2 ] = √ [ 225 - 144 ] = √81 = 9
So....the volume of the cone is
(1/3) pi (radius)^2 * (height) =
(1/3) pi * (12)^2 * 9 =
432 pi units^2
+128475 C, What is the radius, in inches, of a right circular cylinder if the lateral surface area is 24 pi square inches and the volume is 24 pi cubic inches?
If the volume is 24 pi in^3, we have that
24 pi = pi ^ r^2 * h
Solving for h we have that
24 / r^2 = h (1)
And if the lateral surface area is 24 pi in^2......we have that
24 pi = 2 pi * r * h
Sub (1) for h and we have
24 pi = 2 pi * r * (24/r^2 )
1 = 2 / r
r = 2 in
+128475 Here's my attempt at D.....
Surface area of each donut hole covered by Akshaj = 4 pi (10)^2 = 400 pi mm^2
Surface area of each donut hole covered by Theo = 4 pi (8)^2 = 256 pi mm^2
Surface area of each donut hole covered by Niraek = 4 pi (6)^2 = 144 pi mm^2
Ignoring the "pi's" and prime factoring each number we have that
5^2 * 2^4 = 400
2^8 = 256
2^4 * 3^2 = 144
The LCM of these three numbers is :
2^8 * 5^2 * 3^2 = 57600
So.....when they all finish a donut hole at the same time :
Akshaj will have covered 57600 / 400 = 144 donut holes
Theo will have covered 57600 / 256 = 225 donut holes
Niraek will have covered 57600 / 144 = 400 donut holes
EDIT to correct an error!!!!
