The volume of ice-cream in the cone is half the volume of the cone. The cone has a 3cm radius and 6cm height. What is the depth of the ice cream, correct to two decimal places?

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The volume of ice-cream in the cone is half the volume of the cone. The cone has a 3cm radius and 6cm height. What is the depth of the ice cream, correct to two decimal places?

-measurements is in cm

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Answer 1 · 2018-01-28T00:58:55

$h = 4.76$ $h=4.76$ $c m$ $cm$

Explanation:

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$V_{Cone} = π R^{2} \frac{H}{3}$ $V_"Cone"=piR^2H/3$

$R = 3$ $R=3$ $c m$ $cm$

$H = 6$ $H=6$

$V_{Cone} = π {(3)}^{2} (\frac{6}{3}) = 18 π$ $V_"Cone"=pi(3)^2(6/3)=18pi$ $c m^{3}$ $cm^3$

$V_{Ice Cream} = \frac{18 π}{2} = 9 π$ $V_"Ice Cream"=(18pi)/2=9pi$ $c m^{3}$ $cm^3$

$r$ $r$ is the radius of the Ice Cream.

$h$ $h$ is the height (depth) of the Ice Cream.

The right triangles formed by the height of the cone and the two radii $R$ $R$ and $r$ $r$ are similar triangles by $AA$ $"AA"$ theorem. Therefore, the ratio of their corresponding sides are the same:

$\frac{h}{6} = \frac{r}{3}$ $h/6=r/3$

$6 r = 3 h$ $6r=3h$

$r = \frac{3 h}{6} = \frac{h}{2}$ $r=(3h)/6=h/2$

$V_{Ice Cream} = π r^{2} \frac{h}{3}$ $V_"Ice Cream"=pir^2h/3$

$V_{Ice Cream} = π {(\frac{h}{2})}^{2} (\frac{h}{3})$ $V_"Ice Cream"=pi(h/2)^2(h/3)$

$9 π = \frac{h^{3}}{12} π$ $9pi=h^3/12pi$

$9 = \frac{h^{3}}{12}$ $9=h^3/12$

$h^{3} = 108$ $h^3=108$

$h = \sqrt[3]{108} = 4.76$ $h=root(3)108=4.76$ $c m$ $cm$

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The volume of ice-cream in the cone is half the volume of the cone. The cone has a 3cm radius and 6cm height. What is the depth of the ice cream, correct to two decimal places?

-measurements is in cm

1 Answer

Explanation:

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