Question

The volume of a right circular cone varies jointly as the altitude and the square of the radius of the base. If the volume of the cone is 154 cu. in. when its altitude is 12 in. and the radius of the base is 3 1/2 in., find the altitude when the volume of the cone is 77 cu. in. and the radius of the base is 2 1/3 in. Altitude = ? inches

Answer 1

Answer:13.5 inches

Step-by-step explanation:

Here, the volume of a right circular cone varies jointly as the altitude and the square of the radius of the base

So our equation for volume becomes

V = c*h*r^2

where 'h' is altitude and 'r' is radius of base and 'c' is constant

Putting the value of V,h,r,

we get,

154 = c*12*3.5*3.5

c = 22/21

Now we have volume = 77 cu and radius of the base =7/3, so putting the values we get,

77 = 22/21*h*7/3*7/3

or, h= 13.5

Hope it helps!!!

Answer 2

Answer: 13.5 inches.

Step-by-step explanation:

Given : The volume of a right circular cone varies jointly as the altitude and the square of the radius of the base.

V α r² h , where r= radius and h = height.

i.e. V = k r² h    (1), where c is the constant of proportionality.

When the volume of the cone is 154 cu. in. when its altitude is 12 in. and the radius of the base is $3\dfrac{1}{2}\ in.$ .

Put $V= 154$  ,  $r =3\dfrac{1}{2}\ in. =\dfrac{7}{2}\ in.$ and h = 12

in (1) , we get

$154= k(\dfrac{7}{2})^2(12)$

$154= k \dfrac{49}{4}(12)$

$154= k (147)\\\\\Rightarrow\ k=\dfrac{154}{147}=\dfrac{22}{21}$

When the volume of the cone is 77 cu. in. and the radius of the base is $2\dfrac{1}{3}\ in.$

Put V = 77 , $r=2\dfrac{1}{3}=\dfrac{7}{3}\ in.$ and $k=\dfrac{22}{21}$ in (1) , we get

$77=(\dfrac{22}{21})(\dfrac{7}{3})^2h$

$77=(\dfrac{22}{21})(\dfrac{49}{9})h$

$77\times\dfrac{21}{22}\times\dfrac{9}{49}=h\\\\\Rightarrow\ h=\dfrac{27}{2}=13.5\ in.$

Hence, the altitude = 13.5 inches.