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Home » Thus, the shape is an **elliptic paraboloid?** No â squaring led to ellipsoid. - go-checkin.com

Thus, the shape is an elliptic paraboloid? No â squaring led to ellipsoid. - go-checkin.com

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 \boxed{2 + \sqrt{3}}

📌 Question: A historian discovers a triangular tablet with sides 17 cm, 25 cm, and 28 cm. What is the radius of the circumscribed circle around the triangle?

📌 Solution: Use the formula for the circumradius $R$ of a triangle:

📌 R = \frac{abc}{4A}

📌 where $a = 17$, $b = 25$, $c = 28$, and $A$ is the area. First, compute the semi-perimeter:

📌 s = \frac{17 + 25 + 28}{2} = 35

📌 Then compute the area using Heron’s formula:

📌 A = \sqrt{35(35 - 17)(35 - 25)(35 - 28)} = \sqrt{35 \cdot 18 \cdot 10 \cdot 7} = \sqrt{44100} = 210

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rac{1}{k(k+2)} = rac{A}{k} + rac{B}{k+2}.
Multiply both sides by $ k(k+2) $:
= A(k+2) + Bk.
Set $ k = 0 $: $ 1 = A(2) \Rightarrow A = rac{1}{2} $.
Set $ k = -2 $: $ 1 = B(-2) \Rightarrow B = -rac{1}{2} $.
rac{1}{k(k+2)} = rac{1}{2}\left( rac{1}{k} - rac{1}{k+2}
Now substitute into the sum:
\sum_{k=1}^{50} rac{1}{k(k+2)} = rac{1}{2} \sum_{k=1}^{50} \left( r...
This is a telescoping series. Write out the terms:
rac{1}{2} \left[ \left( rac{1}{1} - rac{1}{3}

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