Similar frustums

The upper and lower radii of a frustum of a right circular cone are 8 cm and 32 cm, respectively. If the altitude of the frustum is 10 cm, how far from the bottom base must a cutting plane be made to form two similar frustums?

Correct answer:

x =  6.6667 cm

Step-by-step explanation:

$$\begin{gathered} r=8\text{cm} \\ R=32\text{cm} \\ h=10\text{cm} \\ h=x+y \\ r:y=m:x \\ \frac{r}{y}=\frac{m}{h-y} \\ m=r+y/h\cdot (R-r) \\ r\cdot (h-y)=m\cdot y \\ r\cdot (h-y)=(r+y/h\cdot (R-r))\cdot y \\ 8\cdot (10-y)=(8+y/10\cdot (32-8))\cdot y \\ -2.4y^2-16y+80=0 \\ 2.4y^2+16y-80=0 \\ a=2.4;b=16;c=-80 \\ D=b^2-4ac=16^2-4\cdot 2.4\cdot (-80)=1024 \\ D>0 \\ {y}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-16\pm \sqrt{1024}}{4.8} \\ {y}_{1,2}=-3.333333\pm 6.666667 \\ {y}_1=3.333333333 \\ {y}_2=-10 \\ y=y_1=3.3333=\frac{10}{3}\doteq 3.3333\text{cm} \\ x=h-y=10-3.3333=\frac{20}{3}\doteq 6.6667\text{cm} \\ m=r+y/h\cdot (R-r)=8+3.3333/10\cdot (32-8)=16\text{cm} \\ \text{ Verifying Solution: } \\ {c}_1=r/y=8/3.3333=\frac{12}{5}=2.4 \\ {c}_2=m/x=16/6.6667=\frac{12}{5}=2.4 \\ {c}_1=c_2 \end{gathered}$$

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