Question #14537

2 Answers
P dilip_k
Nov 1, 2016

Let lateral surface area of the cone be #S_l=pirl#
And
Let base surface area of the cone be #S_b=pir^2#

Where r is the radius of circular base and l is slant height.

By the problem we have

#S_l+S_b=2712.96.....(1)#

#S_l-S_b=678.24......(2)#

Adding (1) and (2)

#2S_l=3391.2#

#=>S_1=1695.6#

#=>pixxrxxl=1695.6#

#=>rxxl=1695.6/pi=1695.6/3.14=540...(3)#

Again subtracting (2) from (1)

#=>2S_b=2034.72#

#=>S_b=1017.36#

#pixxr^2=1017.36#

#=>r^2=1017.36/pi=1017.36/3.14=324#

#=>r=18cm.....(4)#

From (3) and (4)

#l=540/18=30#

So height of the cone

#h=sqrt(l^2-r^2)=sqrt(30^2-18^2)=24cm#

Volume of the cone

#V=1/3xxS_bxxh#

#=>V=1/3xx1017.36xx24cm^3=8138.88cm^3#

P dilip_k
Nov 1, 2016

Let lateral surface area of the cone be #S_l=pirl#
And
Let base surface area of the cone be #S_b=pir^2#

Where r is the radius of circular base and l is slant height.

By the problem we have

#S_l+S_b=2712.96.....(1)#

#S_l-S_b=678.24......(2)#

Adding (1) and (2)

#2S_l=3391.2#

#=>S_1=1695.6#

#=>pixxrxxl=1695.6#

#=>rxxl=1695.6/pi=1695.6/3.14=540...(3)#

Again subtracting (2) from (1)

#=>2S_b=2034.72#

#=>S_b=1017.36#

#pixxr^2=1017.36#

#=>r^2=1017.36/pi=1017.36/3.14=324#

#=>r=18cm.....(4)#

From (3) and (4)

#l=540/18=30#

So height of the cone

#h=sqrt(l^2-r^2)=sqrt(30^2-18^2)=24cm#

Volume of the cone

#V=1/3xxS_bxxh#

#=>V=1/3xx1017.36xx24cm^3=8138.88cm^3#

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