Question 119193
Think of {{{y+1}}} as a number. For example if y=1, then {{{y+1=1+1=2}}}. So we can still use this formula



{{{A=(1/2) bh}}} Start with the given formula



{{{A=(1/2) (y+1)(2y)}}} Plug in {{{b=y+1}}} and {{{h=2y}}}




{{{A=(1/2)(2y)(y+1)}}} Rearrange the terms




{{{A=y(y+1)}}} Multiply {{{2y}}} and {{{1/2}}} to get {{{y}}}



{{{A=y^2+y}}} Distribute



So the area of a triangle with base {{{b=y+1}}} and height {{{h=2y}}} is {{{A=y^2+y}}}