Question

A semicircular sheet of metal of diameter $28$ cm is the bent into an open conical cup. Find the depth and the capacity of the cone.

Answer 1

Diameter of the circular sheet is $28\text{cm}.$

Radius $=14\text{cm}$

Circumference of semicircular sheet $=14\pi \text{cm}$

Now the semicircular sheet is bent into an open circular cup.

Slant height of conical cup $=l=$ radius of circular sheet $=14\text{cm}$

Circumference of the base of conical cup = Circumference of semicircle sheet = $\pi r=14\pi$ $(\because R=7\text{cm})$

Depth of conical cup = Height of cone $=h=\sqrt{l^2-r^2}$

$=\sqrt{(14{)}^2-(7{)}^2}$

$=\sqrt{196-49}$

$=\sqrt{147}$

$=7\sqrt{3}\text{cm}$

Capacity of the conical cup $=\frac{1}{3}\pi R^{2}h$

$=\frac{1}{3}\times \frac{22}{7}\times 7^2\times 7\sqrt{3}$

$=\frac{1078\sqrt{3}}{3}{\text{cm}}^3$