Discussion
In a right-angled trapezium \(ABCD\), \(\angle A = 90^{o}\), \(\angle D = 90^{o}\), \(AB = 7a + 1\), \(CD = 3a + 1\) and \(AD = 3a\). If the perimeter of the trapezium is greater than 56 but less than 92, then the range of possible values of \(a\), is
Answer: E
The range is \(3 \lt a \lt 5\).
With right angles at A and D, the parallel sides are AB and CD and AD is the height. Drop a perpendicular from C to AB: it has length \(3a\) and cuts off \(AB - CD = 4a\).
\[ \begin{aligned} BC &= \sqrt{(3a)^{2} + (4a)^{2}} = 5a \\ \text{perimeter} &= (7a + 1) + 5a + (3a + 1) + 3a \\ &= 18a + 2 \end{aligned} \]
\[ \begin{aligned} 56 \lt 18a + 2 &\lt 92 \\ 54 \lt 18a &\lt 90 \\ 3 \lt a &\lt 5 \end{aligned} \]
Check the end points: \(a = 3\) gives a perimeter of exactly 56 and \(a = 5\) exactly 92, both excluded by the strict inequalities.
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