Question

A semicircular sheet of metal of diameter 28 cm is bent into an open conical cup.Find the depth and the capacity of the cup. $(\sqrt{3}=1.73)$

Answer 1

when we bent some water that into open conical cup radius of shut= slant of cone and perimeter (circumference) g shut = circumference g arcular base g cone.

Given Diameter $=17cm$ (shut)

So, $r=\frac{28}{2}=14cm$

and as said above $l=r=14cm$ where $l=$ slant highest

circumference of shut $\pi r=\pi \left(14\right)=\frac{22}{7}\times 14=44cm$

So as said above $44cm=2\pi R$ where $2\pi R$ is circumference of base

$44=2\times \frac{22}{7}+R\Rightarrow R=7m$

In cone $R^2+H^2=l^2$

$\left(7^2\right)+H^2=(14{)}^2$

$H^2=196-49=147cm$

$H^2=147=(49\times 3)cm$

$H=7\sqrt{3}cm$ depth of conical cap

capacity volume of cup $=\frac{1}{3}\times R^{2}H=(\frac{1}{3}\times \frac{22}{7}\times (7{)}^2\times 753)c{m}^2$

$=\frac{1}{3}\times \frac{22}{7}\times 7\times 7\times 753=\frac{1078\sqrt{3}}{3}=622.383c{m}^3$

capacity of cup $=622.383c{m}^3$