Two cone trunks $ T_1 $ and $ T_2 $ have a common base of radius equal to $ 8 $ cm, the other bases being concentric circles. Knowing that the radius of the larger base of $ T_1 $ is equal to $ 15 $ cm, and the volume of $ T_1 $ is three times the volume of $ T_2 $, determine the ratio between the base area of the trunk $ T_2 $ and the trunk $ T_1 $, this ratio being between non-common bases

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1 Answer 1


Volume of trunk $T_1$ is:

$V_1=\frac 12 h(8^2\times 3,14+15^2\times 3,14=A_1)$

And that of trunk $T_2$ is:

$V_2=\frac 12 h(8^2\times 3.14+ 3.14 \times R^2=A_2)$

where $h$ is the height of trunks and $A_1$ and $A_2$ are the areas of trunks $T_1$ and $T_2$ on the top.


$\frac 12 h(8^2\times 3,14+15^2\times 3,14=A_1)=\frac 32 h(8^2\times 3.14+ 3.14 \times R^2=A_2)$

$\rightarrow 2\times 8^2\times 3.14+3A_2=A_1 $

$2\times 8^2\times 3.14\approx 402$

$A_1=15^2\times 3.14\approx 706$

so we have:

$ 402+3A_2=(706=A_1)$

dividing both sides by $706$ we get:

$\frac{3A_2}{A_1}=1-\frac{402}{706}=\frac {304}{706} \Rightarrow \frac {A_2}{A_1}=\frac{304}{3\times 706}\approx 0.144=\frac{144}{1000}$


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