A trapezium in the plane is a quadrilateral in which a pair of...

Without losing generality, assume $a>c$ and $d>b$ and sides $A B \| C D$
$$\begin{aligned} & d^2-x^2=b^2-(c-a+x)^2, \text { Similarly } d^2-x^2=b^2-(a-c-x)^2 \\ & \Rightarrow x=\frac{(a-c)^2+d^2-b^2}{2(a-c)} \Rightarrow x=\frac{(a-c)^2+d^2-b^2}{2(a-c)} \end{aligned}$$ If $x \in(0, d)$, then there will be unique trapezoid $$\begin{aligned} & \Rightarrow \quad \frac{(a-c)^2+d^2-b^2}{2 c(c-a)} \in(0, d) \\ & \Rightarrow \quad(a-c)^2+d^2-b^2-2 d(a-c) < 0 \\ & \Rightarrow \quad(a-c-d)^2-b^2 < 0 \Rightarrow(a-c-d-b)(a-c-d+b) < 0 \\ & \Rightarrow \quad(a-c-d+b)>0 \Rightarrow a+b>c+d \end{aligned}$$ And $a>c, d>b$ Using these inequality, numerate these pairs ( $a, b, c, d$ ) Case I : $a=10 \Rightarrow$ total no. of cases $=16$ Case II : $a=9 \Rightarrow$ total no. of cases $=9$ Case III : $a=8 \Rightarrow$ total no. of cases $=4$ Case (IV) : $a=7 \Rightarrow(7,9,5,10)$ and ( $7,8,5,9$ ) $\Rightarrow 2$ cases $\Rightarrow \quad$ Total $=31$