We have a rectangle that has a width that is 2 inches longer than its height.
We know that the diagonal is 28 inches long.
We can draw this as:
We can use the Pythagorean theorem to relate the measures as:
[tex]\begin{gathered} h^2+w^2=d^2 \\ h^2+(h+2)^2=28^2 \\ h^2+h^2+4h+4=784 \\ 2h^2+4h+4-784=0 \\ 2h^2+4h-780=0 \\ h^2+2h-390=0 \end{gathered}[/tex]The height will come from the solution of this quadratic equation:
[tex]\begin{gathered} h=\frac{-2\pm\sqrt{2^2-4\cdot1\cdot(-390)}}{2\cdot1} \\ \\ h=\frac{-2\pm\sqrt{4+1560}}{2} \\ \\ h=\frac{-2\pm\sqrt{1564}}{2} \\ \\ h=\frac{-2\pm39.547}{2} \\ \\ h=-1\pm19.77 \\ \Rightarrow h_1=-1-19.77=-20.77 \\ \Rightarrow h_2=-1+19.77=18.77 \end{gathered}[/tex]The negative solution does not make sense in the context of the problem, so the solution to this problem is h = 18.8 inches.
Answer: the height is approximately 18.8 inches.
