If x, y, z be the height, curved surface area and volume of a cone respectively, then (3*Pi*z*x^3 -

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25 Dec 2015 Surface area

If x, y, z be the height, curved surface area and volume of a cone respectively, then (3*Pi*z*x^3 - y^2*x^2 + 9z^2)

OPtion

1) 2 Pi
2) 16 Pi^2
3) 4 Pi
4) 0
5) 8 Pi
6) 8
7) 32 Pi^2
8) 4
9) 2 Pi^2
10)None of these

Solution

Let radius r
y = Pi r Sqrt(r^2 + x)
y^2 = Pi^2 r^2 (r^2 + x)
and z = 1/3 Pi r^2 x
Using this and solve
answer is 0

Option 4)

Correct Option: 4 Best Solution (14)

Bryan Tsai   9 years AGO

x = h
y = PI * r * l
z = PI * r^2 * h/3

l^2 = r^2 + h^2

3 PI * z * x^3 - y^2 * x^2 + 9 z^2 =
3 PI * PI * r^2 * h/3 * h^3 - PI ^2 * l^2 * h^2 + 9 PI^2 * r^4 * h^2/9 =
PI^2 * r^2 * h^4 - PI^2 * r^2 * l^2 * h^2 + PI^2 * r^4 * h^2 =
PI^2 * r^2 * h^2 * (r^2 + h^2) - PI^2 * r^2 * h^2 * l^2 =
PI^2 * r^2 * h^2 * l^2 - PI^2 * r^2 * h^2 * l^2 =
0

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ANUP KUMAR BARICK   9 years AGO

height = h,
surface area= pi*r*s + pi*r^2
volume =pi*r^2*h÷3

by applying in main equation , it will be zero

Like? Yes (1) | No   

MRITUNJAY   9 years AGO

put the value of x y and z we get the value
of equation =0
option 4

Like? Yes (2) | No (1)  

NIKHIL JAIN   9 years AGO

csa= pi*r*l= pi*r* (h^2+r^2)^1/2
volume= 1/3 * pi* r^2* h

on solving the above equation, answer = 0

Like? Yes (1) | No   

Vivek Ratan   9 years AGO

csa of cone is pi*r*l where l=(r.(sq)+x(sq))
volume of cone is 1/3*pi*r(sq)*x

substituting the values in above equation and solving we get 0

Like? Yes (1) | No   

sumit kumar   9 years AGO

cone volume z= 1/3*pi*r^2*x [height=x]
surface area y=pi*r*l=pi*r*(x^2+r^2)^1/2
so puting these value in question
3*pi*z*x^3 -y^2*x^2 +9z^2
we get 0
so option 4

Like? Yes (1) | No   

Jaikumar Bhatia   9 years AGO

Height of cone =h=x
Curved surface area of cone=pi*r*sqrt(r^2+x^2)=y
Volume of cone=(1/3)*pi*r^2*x=z
Given
(3pizx^3)-(y^2x^2)+(9z^2)....................................(i)
by putting the values of y and z in equation (i)
={3pi(1/3pir^2x)x^3}-{[pirsqrt(r^2x^2]^2x^2}+{9(1/3pir^2x)^2}
={pi^2r^2x^4}-{(pi^2r^4x^2)+(pi^2r^2x^4)}+{pi^2r^4x^2}
=0
ans is 0
option is (4)

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Anurag Bawankar   9 years AGO

x=h
y=pi r sqrt(r^2+h^2)
z=pi r^2 h/after putting these values all the terms get cancled

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BODLA PAVANI   9 years AGO

x=h,y=pi*r*l,z=1/3(pi*r^2*h and l^2=h^2+r^2. substitute these values in the equation will get zero.

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Jamil Ahmed   9 years AGO

Let r be the radius, then slant height = square root (r^2+x^2)

Therefore, y = Pi*r*square root(r^2+x^2) and z = Pi*r^2*x/3

Now,
3*Pi*z*x^3 - y^2*x^2 + 9z^2
= Pi^2*r^2*x^4 - Pi^2*r^2*(r^2+x^2)*x^2 + Pi^2*r^4*x^2
= Pi^2*r^2*x^4 - Pi^2*r^4*x^2 - Pi^2*r^2*x^4 + Pi^2*r^4*x^2
= 0

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RAHUL AGARWAL   9 years AGO

(Curved surface area of cone) y=pi*r*l, where r & l are radius & slant height of cone.
(volume of cone) z=pi* r^2*x/3 , where x is height of cone.
l^2=r^2+x^2
3*pi*z*x^3-y^2*x^2+9z^2 = pi^2*r^2*x^4-pi^2*r2*l^2*x^2+pi^2*r^4*x2
=pi^2*r^2*x^4-pi^2*r^2*(r^2+x^2)*x^2+pi^2*r^4*x^2
cancel all the terms we get 0
=0 (ans)

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ROHIT KUMAR DUBEY   9 years AGO

as a curve surface area of cone=pi*r*l
where l^2=(r^2+h^2)
volume=1/3 pi*r^2*h
putting all value in given equation
we get result =0

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RENU KUMARI   9 years AGO

by putting 1/3pir^2h in place of volume and pirl where l^2=r^2+h^2 in place of curved surface area we get zero.

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Nandkishor jaiswal   9 years AGO

given that
x=h
y=pi*r*l
z=pi*r^2*h
from solving the equation
3*pi*2*x^3-y^2*x^2+9*z^2=0

Like? Yes (1) | No (1)