Question

plz solve the whole worksheet except ques 1 and 2
solve the remaining
solve all
donot miss anything

Answer 1

3.) h=2 L
   r=r/2
then total surface area=$\pi r^{2} + \pi rl$
=[tex] now--give--the--values \\ \pi * (\frac{r}{2} ) ^{2} + \pi * \frac{r}{2} *2l \\ \\ \ \frac{ \pi r^{2} }{4}+ \pi rl \\ \\ [/tex]$\pi r(l+ \frac{r}{4} )$

4).the volume of cone and cylinder will be same
   volume of the cylinder=πr²h
cylinder's r is= cone's r
let the height is h2
volume of the cone=$\frac{1}{3} \pi r^{2} h _{2}$
$\frac{1}{3} \pi r^{2} h _{2}$=πr²h
cancel pi and r
h=$\frac{1}{3} h _{2}$
ratio between h2 and h is 3:1

10. volume of a cone(v)=[tex] \frac{1}{3} \pi r^{2} h [/tex]
=$\frac{ \pi r^{2} h}{3}$
now r=2r
h=2h
volume=$\frac{1}{3} \pi (2r) ^{2} *2h$
           =[tex] \frac{1}{3} \pi 4 r^{2} *2h \\ \\ =\frac{8 r^{2} h \pi }{3} [/tex]
compare two equations....
so u will get volume=8v

12.)  r=r
volume of cone=$\frac{1}{3} \pi r ^{2} h$
volume of cyliner= \pi r ^{2} H
v=v
[tex] \pi r^{2} H=\frac{ \pi r^{2} h }{3} \\ H= \frac{h}{3} [/tex]
height of the cone: height of the cylinder=1:3