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Home » This is a telescoping sum. Write out the first few terms: - Crosslake

This is a telescoping sum. Write out the first few terms: - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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So, the area of the original hexagon is:
A_1 = \frac{3\sqrt{3}}{2} \times 10^2 = 150\sqrt{3} \text{ cm}^2
For the new hexagon with radius 11 cm:
s' = 11 \text{ cm}
The area of the new hexagon is:
A_2 = \frac{3\sqrt{3}}{2} \times 11^2 = \frac{3\sqrt{3}}{2} \times 121...
\Delta A = A_2 - A_1 = 181.5\sqrt{3} - 150\sqrt{3} = 31.5\sqrt{3} \tex...
Thus, the area of the hexagon increases by:
\boxed{31.5\sqrt{3} \text{ cm}^2}
Question:** A right circular cone has a height of 12 cm and a base rad...

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