Respuesta :
Recall that the area of a triangle is [tex]\frac12bh[/tex] and the area of a parallelogram is [tex]bh[/tex].
The bases, [tex]b[/tex], are equal, but the heights, [tex]h[/tex] are not necessarily. Let us use [tex]h_t[/tex] to denote the height of the triangle and [tex]h_p[/tex] to denote the height of the parallelogram.
The areas of the triangle and the parallelogram are equal: [tex]\frac12bh_t=bh_p[/tex]. Thus [tex]\frac12h_t=h_p[/tex] and [tex]h_t=2h_p[/tex].
That is, the height of the triangle is twice the height of the parallelogram. The answer is c.
The bases, [tex]b[/tex], are equal, but the heights, [tex]h[/tex] are not necessarily. Let us use [tex]h_t[/tex] to denote the height of the triangle and [tex]h_p[/tex] to denote the height of the parallelogram.
The areas of the triangle and the parallelogram are equal: [tex]\frac12bh_t=bh_p[/tex]. Thus [tex]\frac12h_t=h_p[/tex] and [tex]h_t=2h_p[/tex].
That is, the height of the triangle is twice the height of the parallelogram. The answer is c.