Truncated cone

Find the volume and surface area of the truncated cone if r1 = 12 cm, r2 = 5 cm and side s = 10 cm.

Correct answer:

V =  1712.5734 cm³
S =  1064.9999 cm²

Step-by-step explanation:

$$\begin{gathered} r_1=12\text{ cm } \\ {r}_2=5\text{ cm } \\ s=10\text{ cm } \\ {s}^2=h^2+(r_1-r_2{)}^2 \\ h=\sqrt{s^2-(r_1-r_2{)}^2}=\sqrt{10^2-(12-5{)}^2}=\sqrt{51}\text{ cm}\doteq 7.1414\text{ cm } \\ V={\displaystyle \frac{\pi \cdot h}{3}}\cdot (r_1^2+r_1\cdot r_2+r_2^2)={\displaystyle \frac{3.1416\cdot 7.1414}{3}}\cdot (12^2+12\cdot 5+5^2)=1712.5734{\text{cm}}^3 \end{gathered}$$

$S=\pi \cdot r_1^2+\pi \cdot r_2^2+\pi \cdot (r_1+r_2)\cdot \sqrt{h^2+(r_1-r_2{)}^2}=3.1416\cdot 12^2+3.1416\cdot 5^2+3.1416\cdot (12+5)\cdot \sqrt{7.1414^2+(12-5{)}^2}=1064.9999{\text{cm}}^2$

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