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(A) Let the parallel sides of the isosceles trapezium be $a$ and $b$,and the non-parallel sides be $c$. Since it is an isosceles trapezium,the non-par

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$\frac{1}{2}(\sqrt{5}+1)$

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(A) Let the parallel sides of the isosceles trapezium be $a$ and $b$,and the non-parallel sides be $c$. Since it is an isosceles trapezium,the non-parallel sides are equal. The diagonal is $d$. The set of lengths is $\{a, b, c, d\} = \{a, b\}$.Since $a > b$,the sides must be $a, b, a, a$ and the diagonal $d = a$ or $d = b$.Case $1$: The sides are $a, a, a, b$ and the diagonal is $a$. In an isosceles trapezium with parallel sides $a$ and $b$ and non-parallel sides $a$,the diagonal $d$ is given by $d^2 = a^2 + ab - a^2 = ab$ (incorrect geometry).Correct approach: For an isosceles trapezium with parallel sides $a, b$ and non-parallel sides $c$,the diagonal $d$ satisfies $d^2 = c^2 + ab$. Given the set is $\{a, b\}$,we have $c=a$ and $d=a$. Thus $a^2 = a^2 + ab$ is impossible.If $c=b$ and $d=a$,then $a^2 = b^2 + ab$. Dividing by $b^2$,we get $(a/b)^2 - (a/b) - 1 = 0$. Solving for $x = a/b$,$x = \frac{1 + \sqrt{5}}{2}$.

The lengths of the sides and the diagonal of an isosceles trapezium form a two-element set $\{a, b\}$. If $a > b$,then $a / b$ equals

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