The iron roller

The iron roller has a base circumference of 28 π cm. The worker drilled a hole through the top of the roller. After drilling, the given product had a 35% smaller volume than before. The hole's circumference in the base is equal to the height of the roller.

Final Answer:

D₂ =  22.5743 cm
S =  11698.9664 cm²

Step-by-step explanation:

$$\begin{gathered} o_1=28\pi =28\cdot 3.1416\doteq 87.9646\text{cm} \\ r=o_1/(2\pi )=87.9646/(2\cdot 3.1416)=14\text{cm} \\ {D}_1=2\cdot r=2\cdot 14=28\text{cm} \\ {S}_1=\pi \cdot r^2=3.1416\cdot 14^2\doteq 615.7522{\text{cm}}^2 \\ q=100\%-35\%=1-\frac{35}{100}=0.65 \\ {S}_2=q\cdot S_1=0.65\cdot 615.7522\doteq 400.2389{\text{cm}}^2 \\ {r}_2=\sqrt{S_2/\pi }=\sqrt{400.2389/3.1416}\doteq 11.2872\text{cm} \\ {D}_2=2\cdot r_2=2\cdot 11.2872=22.5743\text{cm} \end{gathered}$$

$$\begin{gathered} o_2=2\pi \cdot r_2=2\cdot 3.1416\cdot 11.2872\doteq 70.9193\text{cm} \\ h=o_2=70.9193\doteq 70.9193\text{cm} \\ {S}_3=o_1\cdot h=87.9646\cdot 70.9193\doteq 6238.3895{\text{cm}}^2 \\ {S}_4=o_2\cdot h=70.9193\cdot 70.9193\doteq 5029.5504{\text{cm}}^2 \\ S=S_3+S_4+2\cdot S_1-2\cdot S_2=6238.3895+5029.5504+2\cdot 615.7522-2\cdot 400.2389=11698.9664{\text{cm}}^2 \end{gathered}$$

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