Fill A Cone

The volume of a cone is given by

$V_{cone} = \frac {1}{3} \pi r^2 h$

The volume of the small cone is given as $3 ft^2$ and the height of this cone is $h = 4 ft$ . Plugging in what we know into the formula:

$3 ft^3 = \frac {1}{3} \pi r^2 (4 ft)$

$9 ft^3 = \pi r^2 (4 ft)$

$\frac{9}{4} ft^2 = \pi r^2$

Let us now consider the bigger cone. Since the height has tripled (from $4 ft$ to $12 ft$ ), it follows that the radius has also tripled (to $3r$ ). Thus, if

$\frac{9}{4} ft^2 = \pi r^2$

then

$\pi (3r)^2 = \pi (9r^2) = 9(\pi r^2) = 9(\frac{9}{4} ft^2) = \frac{81}{4} ft^2$

and the volume of the big cone is

$V_{cone} = \frac {1}{3} \pi r^2 h = \frac {1}{3} \pi (3r)^2 (12 ft) = (\frac {1}{3})(\frac{81}{4} ft^2)(12 ft) = 81 ft^3$

and to fill the big cone with water, you will need

$81 ft^3 - 3 ft^3 = \boxed{78 ft^3}$

The ratio of height of the smaller cone to the bigger cone are $1:3$ . Thus, the ratio of radius of smaller cone to bigger cone is $1:3$ too. Hence now we get the ratio of the volume of these two cones, namely

$\frac {\pi(\frac {1}{3}r)^{2}(\frac {1}{3}h)}{\pi r h}=\frac {\frac {1}{27}\pi r h}{\pi r h} = 1 : 27$

This gives us the volume of the bigger cone, that is $27$ times the volume of the smaller cone, $27 \times 3 = 81$ . And the rest water we need to fulfill the bigger cone is simply the difference of volume of bigger cone with the smaller one, that is $81-3=\boxed{78}$

let,k and D be any D be any point on open base circle and water level circle respectively so ,CK and BD are radius of bigger and smaller circle of open base circle and water level circle respectively. also , let CK=R1 and BD = R2 NOW ,in triangle ABD and triangle ACK <ABD =<ACK =90 degree <ADB =<AKC ---------------(as BD is parallel to CK , corresponding angle) so, triangle ABD and triangle ACK are similar to each other thus, AB/AC = BD/CK OR, R2/R1 = 4/12 OR, R2/R1 = 1/3
NOW , volume of smaller cone formed by water / volume of whole body = ( R2/R1)(R2/R1)(H2/H1) 3ft/ volume of whole body = (1/3)(1/3)(1/3) 3ft/ volume of whole body=1/27 or , volume of whole body= 81 ft so remaining volume = (81-3 )ft = 78 ft

AC:AB=(radius of B):(radius of C) By that, we get B:C is 1:3. So the ratio of the volume of the cone to the volume of the water is 27:1 We get that volume of the cone is 81. 81-3=78 So the answer is 78 cubic feet.

Let V be the volume of the cone

let v be the volume of water

let H be the height of cone

let h be the height of water

Apply Similar solids

V/v = H^3/h^3

V/3=12^3/4^3

V=81

x=V-v

x=81-4

x=78

The volume of a cone is given by the formula

V=(1/3)pi(r)^2(h)

Given the partial volume, all we have to do was to get the total the maximum volume the cone can contain. The radius of the partial volume (4 ft.^3) is not given so we solve for it.

3 = (1/3)pi(r)^2(4) r = approximately 0.84 ft.

Using similar triangle ratio of height to radius: We solve for the cone's radius (r)

4 : 0.84 ; 12 : r

r = 2.53

All there was left to do was to plug it in the formula for the volume of a cone. And solve for the difference between the cone's area and the given area.

V = 81 ft.^3 - 3 ft.^3

V = 78 ft. ^3

V cone = 1/3 πr^2 h

tan Ө = perp/base = 12/x= 4/r

3 = 1/3 πr^2 4

r = 3/2√π

x = 36/8√π

V cone =1/3 π ×81/(4√π)×(12 )/3 = 81 ft

Required volume of water to fill the cone = 81 – 3 = 78 ft

The answer is 78. ( 81-3=78)

using similarity of triangles we get the ratios , r/R and l/L i.e 1/3 the required volume is = (pi R2 L) - (pi r2 l) = (pi r2 l)( R2/r2 * L/l - 1) which turns out to be 78ft3

proportional to h^3

if we take lower radius as r1 and base(upper)radius as r2 then we have a relation of this radii to the corresponding heights as AB/AC=r1/r2................(1) given volume is 3 ft3 and AB = 4 ft

we find (r1)2=9/(16*pie)

from relation (1) we have (r2)2=81/(16*pie)

thus volume of cone is (4 pie r2 *2 AC)/3=81

thus valume required =81-3 =78 ft3

At first I thought the sum had insufficient data; later found out that it can be easily solved by similarity theorem

V2=V1(H^3/h^3)

V2=(3)(12^3/4^3)

V2=81ft^3

Volume needed to fill the cone = 81 – 3 = 78 ft^3

(4/12)^3 = 3/x. then x is 81 . then 81-3 =78

In small cone, $V_{small} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 \times 4$ or, $3 = \frac{1}{3} \pi r^2 \times 4$ or, $\pi r^2 = \frac{9}{4}$

Let the radius of big cone be $R$ and that of small cone be $r$ . Now if we assume two triangles, one small and one big, in the two cones, then their height and radii are proportional. $\frac{AC}{AB} = \frac{R}{r}$ $or, \frac{12}{4} = \frac{R}{r}$ $or, R = 3r$

The required volume is a frustum with height $12 - 4 = 8 ft$ and radii $R$ and $r$ . The volume of the frustum of a cone is given by: $V_{frustum} = \frac{ \pi h}{3}(R^2 + Rr + r^2)$ $V_{frustum} = \frac{ \pi 8}{3} ((3r)^2 + 3r \times r + r^2)$ $V_{frustum} = \frac{ \pi 8}{3} (9r^2 + 3r^2 + r^2)$ $V_{frustum} = \frac{ \pi 8}{3} \times 13r^2$ $V_{frustum} = \frac{8}{3} \times 13 \pi r^2$ $V_{frustum} = \frac{8}{3} \times 13 \times \frac{9}{4}$ $\boxed{V_{frustum} = 78 ft}$

Considering the ratio of small to big cone, it is quite easy.

I followed the same as mr ralf anthony.-- vbig con -v small cone = 1/3x22/7x 3r x 3 r x 3 h - 1/3 x 22/7 x r x r x h = 78 cu ft.

We know that volume of cone is 1/3 X3.14X rXrXh , With 4 ft height water volume given is 3 feet cube..Sustituting the values we get radius r =3/2 underroot 3.14. ,with height of 12 feet and 3 feet heights ratio is 1:3, so radius of full cone will be 9/2 underroot 3.14 Now quantity of water required to fill the cone will be { 1/3 X3.14 X9/2underroot 3.14 square X12 - 3 } That is 81-3 = 78 feet cube Ans.

K.K.GARG,Imdia