# Find a partial area of a trapezoid

The formula for the area of a trapezoid is

$$A = \frac{(a+b)}{2}h$$

where a and b are the length of each base and h is the trapezoid's height.

So I want to figure out the area of a portion of the trapezoid. Instead of a height of h, I want to figure it out for a height of g. But this portion of the trapezoid still contains the complete smaller base of the original trapezoid. Let a be the smaller base. Because the height is now g (g < h), the second base, b, will be smaller, let's call that new base, f. So,

$$f = a + \frac{g}{h}b$$

So then the area of this portion of the original trapezoid is

$$A_p = \frac{(a+f)}{2}g = \frac{2a+\frac{g}{h}b}{2}g$$

Is this correct, particularly the formula for calculating $$f$$?

P.S. I've added a drawing.

## 2 Answers

The new base would need to have a length between $$a$$ and $$b$$. In your formula for $$f$$, when $$g = h$$, you have $$f = a + b$$, which is incorrect. Instead, what you require is $$f = a + \frac{g}{h}(b-a),$$ which we can check is:

1. a linear function of $$g$$
2. equal to $$a$$ when $$g = 0$$
3. equal to $$b$$ when $$g = h$$.

Hence the desired trapezoid area is $$A_p = \frac{g}{2} (a+f) = \frac{g}{2}\left(2a + \frac{g}{h}(b-a)\right).$$

As shown in the picture, after dropping the perpendiculars, $$f=a+x+y$$.

Also, $$\dfrac xm=\dfrac yn=\dfrac gh$$. Now according to If $\frac{a}{b} = \frac{c}{d}$ why does $\frac{a+c}{b + d} = \frac{a}{b} = \frac{c}{d}$? we have, $$\dfrac{x+y}{m+n}=\dfrac gh$$. So, the new base is $$a+ \dfrac gh(b-a)$$