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Home » Is This the Trick Chitest Uses? Dont Miss These Hidden Benefits! - Professionaloutdoormedia

Is This the Trick Chitest Uses? Dont Miss These Hidden Benefits! - Professionaloutdoormedia

📅 May 15, 2026 👤 scraface

May 15, 2026

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📌 \boxed{\dfrac{15}{4}}

📌 In a right triangle, the length of one leg is \( a \), the length of the other leg is \( b \), and the length of the hypotenuse is \( c \). If the radius of the inscribed circle is \( r \), find the ratio of the area of the circle to the area of the triangle.

📌 The area of the triangle is given by \( \frac{1}{2}ab \). The semi-perimeter \( s \) of the triangle is \( \frac{a + b + c}{2} \). The radius of the inscribed circle \( r \) can be expressed as \( r = \frac{A}{s} \), where \( A \) is the area of the triangle. Therefore, \( r = \frac{\frac{1}{2}ab}{\frac{a + b + c}{2}} = \frac{ab}{a + b + c} \). The area of the inscribed circle is \( \pi r^2 \). Substituting for \( r \), the area of the circle is \( \pi \left(\frac{ab}{a + b + c}\right)^2 \). The ratio of the area of the circle to the area of the triangle is:

📌 \frac{\pi \left(\frac{ab}{a + b + c}\right)^2}{\frac{1}{2}ab} = \frac{2\pi a^2b^2}{(a + b + c)^2 ab} = \frac{2\pi ab}{(a + b + c)^2}

📌 Thus, the ratio is \(\boxed{\frac{2\pi ab}{(a + b + c)^2}}\).

📌 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}

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