40cmx60cm 83619

The stand on which the posters are stuck has the shape of a cone. It is 2.4m tall. The side of the cone is 2.5 m long. How many 40cmx60cm posters can be stuck on the stand so they do not overlap?

Correct answer:

n =  22
n₂ =  15

Step-by-step explanation:

$$\begin{gathered} h=2.4\text{m} \\ s=2.5\text{m} \\ r=\sqrt{s^2-h^2}=\sqrt{2.5^2-2.4^2}=\frac{7}{10}=0.7\text{m} \\ {S}_1=\pi \cdot r\cdot s=3.1416\cdot 0.7\cdot 2.5\doteq 5.4978{\text{m}}^2 \\ x=40\text{cm}\to \text{m}=40:100\text{m}=0.4\text{m} \\ y=60\text{cm}\to \text{m}=60:100\text{m}=0.6\text{m} \\ {S}_2=x\cdot y=0.4\cdot 0.6=\frac{6}{25}=0.24{\text{m}}^2 \\ {n}_1=S_1/S_2=5.4978/0.24\doteq 22.9074 \\ n=\lfloor n_1\rfloor =\lfloor 22.9074\rfloor =22 \end{gathered}$$

$$\begin{gathered} o=2\pi \cdot r=2\cdot 3.1416\cdot 0.7\doteq 4.3982\text{m} \\ r:h=r_2:(h-y) \\ {r}_2=r\cdot \frac{h-y}{h}=0.7\cdot \frac{2.4-0.6}{2.4}=\frac{21}{40}=0.525\text{m} \\ r:h=r_3:(h-2\cdot y) \\ {r}_3=r\cdot \frac{h-2\cdot y}{h}=0.7\cdot \frac{2.4-2\cdot 0.6}{2.4}=\frac{7}{20}=0.35\text{m} \\ {r}_4=r\cdot \frac{h-3\cdot y}{h}=0.7\cdot \frac{2.4-3\cdot 0.6}{2.4}=\frac{7}{40}=0.175\text{m} \\ {o}_2=2\pi \cdot r_2=2\cdot 3.1416\cdot 0.525\doteq 3.2987\text{m} \\ {o}_3=2\pi \cdot r_3=2\cdot 3.1416\cdot 0.35\doteq 2.1991\text{m} \\ {o}_4=2\pi \cdot r_4=2\cdot 3.1416\cdot 0.175\doteq 1.0996\text{m} \\ {N}_2=\lfloor o_2/x\rfloor =\lfloor 3.2987/0.4\rfloor =8 \\ {N}_3=\lfloor o_3/x\rfloor =\lfloor 2.1991/0.4\rfloor =5 \\ {N}_4=\lfloor o_4/x\rfloor =\lfloor 1.0996/0.4\rfloor =2 \\ {n}_2=N_2+N_3+N_4=8+5+2=15 \end{gathered}$$

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