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Home » We know $ an 60^\circ = \sqrt3 $, so: - AMAZONAWS

We know $ an 60^\circ = \sqrt3 $, so: - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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A_1 = \frac{\sqrt{3}}{4} \cdot 12^2 = \frac{\sqrt{3}}{4} \cdot 144 = 3...
New side length: $ 12 + 2 = 14 $ cm. New area:
A_2 = \frac{\sqrt{3}}{4} \cdot 14^2 = \frac{\sqrt{3}}{4} \cdot 196 = 4...
\Delta A = A_2 - A_1 = 49\sqrt{3} - 36\sqrt{3} = 13\sqrt{3}
Thus, the area increases by $ \boxed{13\sqrt{3}} $ square centimeters.
Question: A cave formation is modeled as a right triangle with a hypot...
Solution: Let the hypotenuse be $ z = 25 $ m, and the inradius be $ c...
Also, the inradius $ r $ of a right triangle is:
Given $ r = 5 $, we have:
a + b - 25 = 10 \Rightarrow a + b = 35

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