Question 859379
When you are having trouble with this, it helps when you draw it.
Let's call the height of the parallelogram p and the one of the triangle t.


Both of their bases are the same length. Let's call that B.


The area of a parallelogram is base*height, and of the triangle is HALF of its own (base*height).


The two areas are the same, and so are the bases, and what is different are the heights. So you set up:


{{{B*p = B*t/2}}}
As B can be cancelled (same value on numerators on both sides, you have this:


{{{p = t/2}}}
So we can say that - height of the parallelogram is half of the one of triangle, or, if we multiply the equation with 2
{{{t = 2p}}}
Height of the triangle is double the height of the parallelogram.