The surface

The surface of a truncated rotating cone with side s = 13 cm is S = 510π cm². Find the radii of the bases when their difference in lengths is 10cm.

Correct answer:

r₁ =  15 cm
r₂ =  5 cm

Step-by-step explanation:

$$\begin{gathered} s=13\text{cm} \\ S=510\pi =510\cdot 3.1416\doteq 1602.2123{\text{cm}}^2 \\ S=\pi \cdot r_1^2+\pi \cdot r_2^2+\pi \cdot (r_1+r_2)\cdot s \\ {r}_1=r_2+10 \\ S=\pi \cdot r_1^2+\pi \cdot (r_1-10{)}^2+\pi \cdot (r_1+(r_1-10))\cdot s \\ S=\pi \cdot x^2+\pi \cdot (x-10{)}^2+\pi \cdot (x+(x-10))\cdot s \\ 1602.2122533308=\pi \cdot x^2+\pi \cdot (x-10{)}^2+\pi \cdot (x+(x-10))\cdot 13 \\ -6.2831853071796x^2-18.85x+1696.46=0 \\ 6.2831853071796x^2+18.85x-1696.46=0 \\ a=6.283185;b=18.85;c=-1696.46 \\ D=b^2-4ac=18.85^2-4\cdot 6.283185\cdot (-1696.46)=42991.996771145 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-18.85\pm \sqrt{42992}}{12.566371} \\ {x}_{1,2}=-1.5\pm 16.5 \\ {x}_1=15 \\ {x}_2=-18 \\ {r}_1,r_2>0 \\ {r}_1=x_1=15=15\text{cm} \end{gathered}$$

Our quadratic equation calculator calculates it.

$$\begin{gathered} r_2=r_1-10=15-10=5=5\text{cm} \\ \text{ Verifying Solution: } \\ {S}_2=\pi \cdot r_1^2+\pi \cdot r_2^2+\pi \cdot (r_1+r_2)\cdot s=3.1416\cdot 15^2+3.1416\cdot 5^2+3.1416\cdot (15+5)\cdot 13\doteq 1602.2123{\text{cm}}^2 \\ {S}_2=S \end{gathered}$$

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