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.

enter image description here


2 Answers 2


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, $f=a+x+y$.

enter image description here

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 $a+ \dfrac gh(b-a)$


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