Question

An iron spherical ball of volume 232848 cm³ has been melted and converted into a cone with the vertical angle of 120° . Find its height and base.

Answer 1

v= πr²h/3 
r = htan60° = √3h 
v = π(3h²)h/3 = πh³ 
given that the volume of material present in the cone = 232848 cm
v = πh³ = 232848 
h = ³√(23248/π) ≈ 42.0 cm

Answer 2

Given :

The volume of iron spherical ball = 232848 cubic cm

The spherical ball is melted and re-casted into cone

The vertical angle of cone = 120°

To Find :

height base of cone

Solution :

Let The height of cone = h cm

Let The radius of cone = r cm

Let The base Area of cone = A cm²

Now,

As the vertical angle = 60°

From figure

Tan$\dfrac{120^{\circ}}{2}$  =  $\dfrac{r}{h}$

where r is the radius and h is the height

So, Tan 60° = $\dfrac{r}{h}$

Or,  $\sqrt{3}$ = $\dfrac{r}{h}$

So, radius = r = $\sqrt{3}$ h cm

Again

The spherical ball is converted into cone,

So, Volume of cone = volume of spherical ball'

i.e  $\dfrac{1}{3}$ × π × radius² × height = 232848 cm³

Putting the value of r into cone volume

So,    $\dfrac{1}{3}$ × π × ($\sqrt{3}$ h cm)² × h  = 232848 cm³

Or,    $\dfrac{1}{3}$ × π × 3  × h³  = 232848

Or,    π × h³  = 232848

∴             h³ = $\dfrac{232848}{\pi }$

or,           h³ = $\dfrac{232848}{3.14 }$

i.e            h³ = 74155.4 cm³

So,          h = $\sqrt[3]{74155.4}$

Or, Height of cone = h = 42.01 ≈ 42 cm

Now,

 Radius of base of cone = r = $\sqrt{3}$ h cm

i.e  radius = r = 42$\sqrt{3}$  cm

So, base area of cone = A = π × radius²

i.e      A = 3.14 × ( 42$\sqrt{3}$  cm) ²

Or,  Area = 16616.8 sq cm

Hence ,

The height of the cone is 42 cm  .

And The base area of cone is 16616.8 sq cm  Answer