Let $$𝐴𝐵𝐶𝐷$$ be a quadrilateral in the $$xy-plane$$ with $$𝐴𝐵$$ parallel to $$𝐶𝐷$$ and $$𝐴𝐷 = 𝐵𝐶$$. Suppose $$𝐴 = (0, 0)$$, $$𝐵 = (10, 0)$$, $$𝐶 = (8, 5)$$ and $$𝐷 = (𝑎, 𝑏)$$. Determine the value of $$a^{2}b$$.

Coordinates of the given vertices are
$$A(0,0),\; B(10,0),\; C(8,5),\; D(a,b).$$

1. Condition $$AB \parallel CD$$:
  • The slope of $$AB$$ is $$0$$ (horizontal line).
  • Therefore the slope of $$CD$$ must also be $$0$$, which gives
$$b-5 = 0 \;\Longrightarrow\; b = 5.$$

2. Condition $$AD = BC$$:
  • Compute $$BC$$:
$$BC = \sqrt{(8-10)^{2} + (5-0)^{2}} = \sqrt{(-2)^{2}+25} = \sqrt{29}.$$
  • Write $$AD$$ in terms of $$a$$ and $$b$$:
$$AD = \sqrt{(a-0)^{2} + (b-0)^{2}} = \sqrt{a^{2}+b^{2}}.$$
  • Equate the two lengths:
$$\sqrt{a^{2}+b^{2}} = \sqrt{29}\;\Longrightarrow\; a^{2}+b^{2}=29.$$

3. Substitute $$b = 5$$ into the length equation:
$$a^{2}+25 = 29 \;\Longrightarrow\; a^{2}=4 \;\Longrightarrow\; a = \pm 2.$$

4. Required value $$a^{2}b$$:
$$a^{2}b = 4 \times 5 = 20.$$
(The result is the same for $$a=2$$ or $$a=-2$$ because $$a^{2}$$ is positive.)

Hence, the value of $$a^{2}b$$ is 20.