We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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)

Guest May 16, 2018

#1**+1 **

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

CPhill May 16, 2018

#2**+1 **

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

CPhill May 16, 2018

#3**+1 **

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

CPhill May 16, 2018

#4**+2 **

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!!!!

CPhill May 16, 2018