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A = \frac{\sqrt{3}}{4} s^2
Given $ \frac{\sqrt{3}}{4} s^2 = 36\sqrt{3} $, divide both sides by $ \sqrt{3} $:
\frac{1}{4} s^2 = 36 \Rightarrow s^2 = 144 \Rightarrow s = 12 \text{ cm}
New side length: $ 12 - 4 = 8 $ cm.
New area:
\frac{\sqrt{3}}{4} \cdot 8^2 = \frac{\sqrt{3}}{4} \cdot 64 = 16\sqrt{3}
Decrease in area:
36\sqrt{3} - 16\sqrt{3} = 20\sqrt{3}
Thus, the area decreases by $ \boxed{20\sqrt{3}} $.
Question: A triangle has side lengths 13 cm, 14 cm, and 15 cm. Find the length of the shortest altitude in centimeters.