In a quadrilateral ABCD, \(\angle ABC = 90^\circ\) and \(\angle ACD = 90^\circ\). Let \(AB = p\) units, \(BC = q\) units, \(AC = r\) units, \(CD = s\) units, \(AD = t\) units, where \(p < q < r < s < t < 15\). If p, q, r, s and t are integers, then what is the area of the quadrilateral?
36 square units
Since \(\angle ABC = 90^\circ\), triangle ABC gives \(p^2+q^2=r^2\), and since \(\angle ACD = 90^\circ\), triangle ACD gives \(r^2+s^2=t^2\). With \(p < q < r < s < t < 15\) all integers, the only consistent choice is the Pythagorean triples \((3,4,5)\) and \((5,12,13)\), giving \(p=3, q=4, r=5, s=12, t=13\). Area of ABCD \(= \tfrac12 pq + \tfrac12 rs = \tfrac12(3)(4) + \tfrac12(5)(12) = 6+30 = 36\) square units.
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