Slope of a Side of a Trapezium

The area of a trapezium is $A = \frac{1}{2} h (b_1 + b_2)$ , where $h$ is its height and $b_1$ and $b_2$ are the lengths of two opposite sides. In our problem we can use $h=3$ , so we get the equation $\frac{417}{16} = \frac{3}{2}(b_1+b_2)$ . The side lengths $b_1$ and $b_2$ can be found by plugging in $3$ and $6$ into the equation of the line forming the fourth side: $b_1 = 3m + \frac{5}{2}$ and $b_2 = 6m + \frac{5}{2}$ .

Hence, we solve the equation $\frac{417}{16} = \frac{3}{2}(3m + \frac{5}{2} + 6m + \frac{5}{2})$ for $m$ . Simplifying, we get $\frac{417}{16} = \frac{3}{2}(9m+5)$ and $\frac{417}{16} = \frac{27}{2} m + \frac{15}{2}$ . Mulitplying by $16$ gives us $417 = 216 m + 120$ , and so $m = \frac{297}{216} = \frac{11}{8}$ and $a+b = 19$ .