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2015-03-06T12:00:13+05:30

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3.) h=2 L
   r=r/2
then total surface area= \pi  r^{2} + \pi rl
=
now--give--the--values \\ 
 \pi * (\frac{r}{2} ) ^{2} + \pi * \frac{r}{2} *2l \\  \\ 
 \ \frac{ \pi r^{2} }{4}+ \pi rl \\  \\ 
 \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)= \frac{1}{3}  \pi  r^{2} h

= \frac{ \pi  r^{2} h}{3}
now r=2r
h=2h
volume= \frac{1}{3}  \pi (2r) ^{2} *2h
           = \frac{1}{3}  \pi 4 r^{2} *2h \\  \\ 
=\frac{8 r^{2} h \pi  }{3}
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
 \pi  r^{2} H=\frac{ \pi  r^{2} h }{3}  \\ 
H= \frac{h}{3}
height of the cone: height of the cylinder=1:3


1 5 1
8th qn is not clear..
7th qn is not clear...
i can't see the edge of the qn paper..it is not clear..wt can i do?
ques 5 to 9 and ques 11
are not answered