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Home » From Workbench to Fame: The Incredible Story of the Charisma Carpenter You Need to See! - AMAZONAWS

From Workbench to Fame: The Incredible Story of the Charisma Carpenter You Need to See! - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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📌 A museum curator wants to create a digital display showing a model of an ancient equilateral triangle with side length \( s \). The model includes an inscribed circle. Determine the ratio of the area of the model triangle to the area of the inscribed circle.

📌 The area \( A \) of an equilateral triangle with side length \( s \) is given by \( A = \frac{\sqrt{3}}{4} s^2 \). The radius \( r \) of the inscribed circle in an equilateral triangle is \( r = \frac{s\sqrt{3}}{6} \). The area of the circle is \( \pi r^2 = \pi \left(\frac{s\sqrt{3}}{6}\right)^2 = \frac{\pi s^2 \cdot 3}{36} = \frac{\pi s^2}{12} \). The ratio of the area of the triangle to the area of the inscribed circle is:

📌 \frac{\frac{\sqrt{3}}{4} s^2}{\frac{\pi s^2}{12}} = \frac{\sqrt{3}}{4} \cdot \frac{12}{\pi} = \frac{3\sqrt{3}}{\pi}

📌 Thus, the ratio is \(\boxed{\frac{3\sqrt{3}}{\pi}}\).

📌 An equilateral triangle has a side length of \( s \) and its area is reduced by removing a smaller equilateral triangle from one corner, such that the side of the smaller triangle is \( \frac{s}{2} \). Determine the percentage decrease in the area of the triangle.

📌 The area of the original equilateral triangle is \( A_1 = \frac{\sqrt{3}}{4} s^2 \). The area of the smaller equilateral triangle with side \( \frac{s}{2} \) is \( A_2 = \frac{\sqrt{3}}{4} \left(\frac{s}{2}\right)^2 = \frac{\sqrt{3}}{4} \cdot \frac{s^2}{4} = \frac{\sqrt{3}}{16} s^2 \). The decrease in area is \( \Delta A = A_1 - A_2 = \frac{\sqrt{3}}{4} s^2 - \frac{\sqrt{3}}{16} s^2 = \left(\frac{4}{16} - \frac{1}{16}\right)\sqrt{3} s^2 = \frac{3\sqrt{3}}{16} s^2 \). The percentage decrease is:

📌 \frac{\Delta A}{A_1} \times 100 = \frac{\frac{3\sqrt{3}}{16} s^2}{\frac{\sqrt{3}}{4} s^2} \times 100 = \frac{3}{4} \times 100 = 75\%

📌 Thus, the percentage decrease in area is \(\boxed{75\%}\).

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