# Determine the ratio between the base area of the trunk $T_2$ and the trunk $T_1$

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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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.

$$V_1=3V_2$$

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

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

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

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

so we have:

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

dividing both sides by $$706$$ we get:

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