If the area of a triangle with base X is equal to the area of a square with side X, then the altitude of triangle is
(a)$\displaystyle X/2$
(b)X
(c)2x
(d)3x
Area of a triangle is equal to:
$\displaystyle A = \frac{1}{2} \cdot \text{base} \cdot \text{height}$
Height is also known as altitude. Area of a square is:
$\displaystyle A = \text{side} \cdot \text{side}$
The problem says that:
$\displaystyle A_\text{triangle} = A_\text{square}$
This info is enough for you to find the answer.
What we are doing is this:
$\displaystyle \tfrac{1}{2}bh=s^2$
Since the base of the triangle is x, the side of the square is x, and we want these two areas to be the same, we see that we have the equation:
$\displaystyle \tfrac{1}{2}x\cdot h=x^2$
Solving for h, we get $\displaystyle h=2\frac{x^2}{x}=\color{red}\boxed{2x}$
Does this make sense?
--Chris
EDIT: I see...you just won't let Jhevon and I be called Flash...you want to be a part too, Chops...don't ya??