Question

hemispherical vessel of radius 14 cm is fully filled with the with the sand.This sand is poured on the level ground. The heap of the sand forms a cone shape of height 7 cm. calculate the area of the ground occupied by the circular base of the heap of the sand

Answer 1

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Answer 2

Answer:

the area of the ground occupied by the circular base of the heap of the sand$=2464\text{ cm}^{2}$

Step-by-step explanation:

Given,

Radius of the hemisphere vessel $R=14\text{ cm}$

The volume of the hemisphere vessel

$V_{1}=\frac{2}{3}\pi\times{R}^{3}\\\\\Rightarrow{V}_{1}=\frac{2}{3}\times\frac{22}{7}\times14^{3}$

Now,

Consider the radius of the cone base $=r$

Height of the cone$h=7\text{ cm}$

therefore the volume of the cone

$V_{2}=\frac{1}{3}\pi\times{r}^{2}\times{h}\\\\\Rightarrow{V_{2}}=\frac{1}{3}\times\frac{22}{7}\times{r}^{2}\times{7}$

As per the problem

$V_{1}=V_{2}\\\\\Rightarrow\frac{2}{3}\times\frac{22}{7}\times14^{3}=\frac{1}{3}\times\frac{22}{7}\times{r}^{2}\times{7}\\\\\Rightarrow{2}\times14^{3}={r}^{2}\times{7}\\\\\Rightarrow{r}^{2}=\frac{2\times14^{3}}{7}\\\\\Rightarrow{r}^{2}=784\\\\\Rightarrow{r}=\sqrt{784}\\\\\Rightarrow{r}=28\text{ cm}$

the area of the ground occupied by the circular base of the heap of the sand$A=\pi\times{r}^{2}=\frac{22}{7}\times28^{2}=2464\text{ cm}^{2}$