Question

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

Answer 1

the radius of circular sheet will be slant height of cone.
so,
slant height =14 m.
pie*r 1= circumference of base of cone.
22/7* 14 m=2*22/7*r
r=7 m.
height or depth of cone= √(l*l-r*r)
= √(14×14−7×7)m
=12.12 m.

capacity of cup= 1/ 3*22/7*7*7*12.12 cubic m
=

Answer 2

$\underline\mathfrak\pink{Answer:}$

When the semi-circular sheet is bent into an open conical cup, the radius of the sheet becomes the slant height of the cup and the circumference of the sheet becomes the circumference of the base of the cone.

•°• $\sf\underline{l=Slant\:height\:of\:the\:conical\:cup=14cm}$

Let r cm be the radius of and h cm the height (depth) of the conical cup, Then,

• Circumference of the base of the conical cup = circumference of the sheet

$\Rightarrow$$\sf{2πr=π×14}$

$\Rightarrow$$\sf\underline{r=7cm}$

Now, $\sf{l^2=r^2+h^2}$

$\Rightarrow$$\sf{h=}$$\sf\sqrt{l^2-r^2}$

$\Rightarrow$ $\sf{h=}$$\sf\sqrt{14^2-7^2}=7\sqrt3$cm

$\Rightarrow$ $\sf{(h=7×1.732)cm=12.12cm}$

$\mathfrak\purple{depth\:of\:the\:cup=12.12cm}$

Now, Capacity of cup = volume of cup

Capacity of cup = $\sf\frac{1}{3}πr^2h\:cm^3$

=== Capacity of the cup = 1/3× 22/7 × 7 × 7 × 12.12 cm³ = 622.26 cm³