Solution
It is given that the width of a rectangle is 2 inches longer than the height.
If the height is h inches,
The width of the rectangle is (2 + h) inches
If the diagonal measurement is 58 inches.
Using Pythagoras theorem, the relationship between the three sides is given as
[tex]h^2+(2+h)^2=58^2[/tex]
Expanding the brackets;
[tex](2+h)^2=(2+h)(2+h)=2(2+h)+h(2+h)=4+2h+2h+h^2=4+4h+h^2[/tex][tex]\begin{gathered} \Rightarrow h^2+4+4h+h^2=58 \\ \\ \Rightarrow2h^2+4h=58-4 \\ \\ \Rightarrow2h^2+4h=54 \\ \\ \Rightarrow h^2+2h-27=0 \end{gathered}[/tex]
Using the quadratic formula,
[tex]\begin{gathered} h=\frac{-2\pm\sqrt{2^2-4(1)(-27)}}{2} \\ \\ \Rightarrow h=\frac{-2\pm\sqrt{4+108}}{2} \\ \\ \Rightarrow h=\frac{-2\pm4\sqrt{7}}{2} \\ \\ \Rightarrow h=-1\pm2\sqrt{7} \\ \\ \Rightarrow h\approx4.3\text{ inches} \end{gathered}[/tex]
Hence, the height is 4.3 inches
