Her work is shown below. V = two-thirds + 54. V = two-thirds + StartFraction 162 Over 3 EndFraction. V = StartFraction 164 Over 3 EndFraction meter cubed. What is Amie's error?

Answer :

The question is incomplete. The complete question is :

A sphere and a cylinder have the same radius and height. The volume of the cylinder is 54 meters cubed. Amie found the volume of the sphere. A sphere with height h and radius r. A cylinder with height h and radius r. Her work is shown below. V = two-thirds + 54. V = two-thirds + StartFraction 162 Over 3 EndFraction. V = StartFraction 164 Over 3 EndFraction meter cubed. What is Amie's error?

Solution :

Amie should have multiplied 54 by two-thirds.

Given that sphere and the cylinder have the same radius as well as same height. The volume of the cylinder is 54 inch cube.

Volume of the sphere = [tex]$\frac{4}{3}\pi r^3$[/tex]      .......... (i)

Volume of cylinder = [tex]$\pi r^2h$[/tex]   ....................(ii)

Therefore, the ratio of sphere volume to cylinder volume is,

[tex]$\frac{4}{3}\pi r^3 : \pi r^2 h$[/tex]   ......(iii)

Divide both the sides by [tex]$\pi r^2$[/tex] , we get

[tex]$\frac{4}{3} \ r : h$[/tex]          ..........(iv)

We know that the height of the sphere = diameter of the sphere

The diameter of the sphere is D = 2r

Also the sphere height = cylinder height

So, the height of the cylinder = 2r

Therefore, substituting the height of the cylinder as 2r that is represented as h in equation (iv) is given by :

[tex]$\frac{4}{3} \ r : 2r$[/tex]        .............(v)

Now dividing both the sides by 2r, we get

[tex]$\frac{2}{3} : 1$[/tex]   ..................(vi)

Thus for equation (vi), we see, sphere volume = [tex]$\frac{2}{3}$[/tex]  of the cylinder volume

∴  sphere volume = [tex]$\frac{2}{3}\times 54$[/tex]

                              = 36 meter cube

Thus, Amie should have multiplied 54 by [tex]$2/3$[/tex] .  

Other Questions