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📅 May 1, 2026 👤 scraface

May 01, 2026

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📌 220\pi r = \frac{40}{r^2} \Rightarrow r^3 = \frac{40}{220\pi} = \frac{2}{11\pi}

📌 \Rightarrow r = \left( \frac{2}{11\pi} \right)^{1/3}

📌 Compute numerically: \( \frac{2}{11\pi} \approx \frac{2}{34.5575} \approx 0.05796 \), so \( r \approx (0.05796)^{1/3} \approx 0.387 \) m

📌 Then \( h = \frac{0.5}{\pi (0.387)^2} \approx \frac{0.5}{0.470} \approx 1.064 \) m

📌 Now compute areas:

📌 Lateral area: \( 2\pi r h \approx 2\pi \cdot 0.387 \cdot 1.064 \approx 2.58 \) m²

📌 Top and bottom: \( 2\pi r^2 \approx 2\pi \cdot 0.150 \approx 0.942 \) m²

📌 Cost: \( 40 \cdot 2.58 + 55 \cdot 0.942 \approx 103.2 + 51.81 = 155.01 \)

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The area of the circular sector is:
A_{ ext{sector}} = rac{1}{2} r^2 heta = rac{1}{2} \cdot 16 \cdot r...
Area of triangle \(OAB\):
A_{ ext{triangle}} = rac{1}{2} ab \sin heta = rac{1}{2} (2\sqrt{6})...
ight) = rac{1}{2} \cdot 24 \cdot rac{\sqrt{3}}{2} = 6\sqrt{3}
Area of segment (the smaller region bounded by the chord and arc) is:
A = A_{ ext{sector}} - A_{ ext{triangle}} = rac{16\pi}{3} - 6\sqrt{3}
But since the line \(y = x + 2\) cuts the circle from top-right to bot...
oxed{rac{16\pi}{3} - 6\sqrt{3}}
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