The formula of the area of a triangle with the height h and the base b: [tex]A_\Delta=\dfrac{bh}{2}[/tex] We have: [tex]A_\Delta=221.16\ in^2\\h=29.1\ in\\b=?[/tex] Substitute: [tex]\dfrac{29.1b}{2}=221.16\\\\14.55b=221.16\ \ \ |:14.55\\\\b=15.2\ in[/tex]
