For a right triangle,
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- Using \( A = c \cdot s \), and \( s = \frac{a + b + z}{2} \), but for a right triangle, it can be shown that:
- \frac{\text{Area of circle}}{\text{Area of triangle}} = \frac{\pi c^2}{c \cdot s} = \frac{\pi c}{s}.
- But we aim to express in terms of \( z \) and \( c \) only. However, without loss of generality, use a standard identity:
- c = \frac{a + b - z}{2}, \quad A = \frac{1}{2}ab = \frac{(a + b)^2 - z^2}{4} \cdot \text{wait — better: }
- Alternatively, recall that \( A = r s = c s \), so:
- \frac{\pi c^2}{A} = \frac{\pi c^2}{c s} = \frac{\pi c}{s}.
