Question 420940

given:

a base {{{b}}} of {{{9 in}}} and a height {{{h}}} of {{{4 in}}}

the area of the triangle {{{A=b*h/2}}}

{{{A=(9in)*(4in)/2}}}

{{{A=(9in)*(cross(4)2in)/cross(2)1}}}

{{{A=(9in)*(2in)}}}

{{{A=18(in)^2}}}