What is the area of a triangle with a base of 4v^3 and a height of vu^2?

 The area of the triangle is:     " 2v⁴u² square units" ;
                                            or, write as:  " 2v⁴u² units² " .

Note:  The formula/equation for the area, "A", of a triangle is:

             A = (1/2) * b * h  ;   or:  A = (b * h) / 2 ; 

  in which:   "A" = "area" ; 
                   "b" = "base length" ;
                   "h" = "[perpendicular] height" ; 

We are asked to solve for the area, "A" .

We are given:  "b = 4v³ " ;  "h = vu² " ;

So;  to find the area, "A", of the triangle, we plug these given values into the formula; and solve/ simplify:

     A = (1/2) * b * h ; 

     A = (1/2) * (4v³) * (vu²) ; 

     A = 2v³ * vu² ;

     A = 2v⁴u²  square units.

 The area of the triangle is:     " 2v⁴u² square units" ;
                                            or, write as:  " 2v⁴u² units² " .
 

Sorry can't answer that question