■ GDC allowed ■..perimeter: \(a + b + c = 90\;\;\; \Rightarrow \;\;\;a + b = 90 - c\)Pythagorean theorem: \({a^2} + {b^2} = {c^2}\)similar triangles: \(\dfrac{c}{b} = \dfrac{a}{{11.25}}\;\;\; \Rightarrow \;\;\;ab = 11.25c\)Consider the identity \({\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\)Substituting gives \({\left( {90 - c} \right)^2} = {c^2} + 2\left( {11.25c} \right)\;\;\;\; \Rightarrow \;\;\;\;{c^2} - 180c + 8100 = {c^2} + 22.5c\)\(202.5c = 8100\;\;\;\; \Rightarrow \;\;\;\;c = \dfrac{{8100}}{{202.5}} = 40\)\(ab = 11.25\left( {40} \right)\;\;\;\; \Rightarrow \;\;\;\;ab = 450\;\;\;\; \Rightarrow \;\;\;\;b = \dfrac{{450}}{a}\)Substituting into perimeter and then multiplying by \(a\) gives: \(a + \dfrac{{450}}{a} + 40 = 90\;\;\;\; \Rightarrow \;\;\;\;{a^2} - 50a + 450 = 0\)\(a = \dfrac{{ - \left( { - 50} \right) \pm \sqrt {{{\left( { - 50} \right)}^2} - 4\left( 1 \right)\left( {450} \right)} }}{{2\left( 1 \right)}} = \dfrac{{50 \pm \sqrt {2500 - 1800} }}{2} = \dfrac{{50 \pm \sqrt {700} }}{2} = \dfrac{{50 \pm \sqrt {100} \sqrt 7 }}{2}\)Hence, \(a = 25 \pm 5\sqrt 7 \)Given that \(a < b\) then \(a = 25 - 5\sqrt 7 \) and \(b = 25 + 5\sqrt 7 \).■ Therefore, the exact lengths of \(a\), \(b\) and \(c\) are \(a = 25 - 5\sqrt 7 \) cm, \(b = 25 + 5\sqrt 7 \) cm,...