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- s' = 12 + 2 = 14 \text{ cm}
- A' = \frac{\sqrt{3}}{4} (14)^2 = \frac{\sqrt{3}}{4} \cdot 196 = 49\sqrt{3} \text{ cm}^2
- \Delta A = A' - A = 49\sqrt{3} - 36\sqrt{3} = 13\sqrt{3} \text{ cm}^2
- Solution: The equation $ |x| + |y| = 4 $ describes a diamond (or rhombus) centered at the origin. To find the vertices, consider the four cases for absolute values:
- $ x \geq 0, y \geq 0 $: $ x + y = 4 $, intercepts at $ (4, 0) $, $ (0, 4) $
- $ x \leq 0, y \geq 0 $: $ -x + y = 4 $, intercepts at $ (-4, 0) $, $ (0