Trapezoid problem

Let $BC=a$ and $AD=b$ . If we label the four horizontal division lines be $B_nC_n$ , where $n=1,2,3,4$ , and the length $B_nC_n = a_n$ . Then $a_n = a + \dfrac {n(b-a)}5$ . Let the height of the trapezoid be $h$ . Then we have:

$\begin{aligned} 2\left(\frac {\frac a5 + \frac {a_1}5}2 \times \frac h5 \right) + \left(\frac {\frac {a_1}5 + \frac {a_2}5}2 \times \frac h5 \right) & = 35 \\ \frac {(9a+b)h}{125} + \frac {(7a+3b)h}{250} & = 35 \\ \implies (5a+b)h & = 1750 & ...(1) \end{aligned}$

Similarly,

$\begin{aligned} \left(\frac {\frac {a_3}5 + \frac {a_4}5}2 \times \frac h5 \right) + 2\left(\frac {\frac {a_4}5 + \frac b5}2 \times \frac h5 \right) & = 55 \\ \frac {(3a+7b)h}{250} + \frac {(a+9b)h}{125} & = 55 \\ \implies (a+5b)h & = 2750 & ...(2) \end{aligned}$

From $(1)+(2): \ \ 6(a+b) h = 4500$ , then the area of the trapezoid $\dfrac {(a+b)h}2 = \dfrac {4500}{12} = \boxed{375}$ .

Let $|\overline {BC}|=a$ , $|\overline {AD}|=b$ and the height of the trapezoid be $h$ . Then $2\times \dfrac{1}{2}\times (\dfrac{a}{5}+\dfrac{4a+b}{25})\times \dfrac{h}{5}+\dfrac{1}{2}\times (\dfrac{4a+b}{25}+\dfrac{3a+2b}{25})\times \dfrac{h}{5}=35$ or $h=\dfrac{250\times {35}}{5(5a+b)}$ . And $2\times \dfrac{1}{2}\times (\dfrac{b}{5}+\dfrac{a+4b}{25})\times \dfrac{h}{5}+\dfrac{1}{2}\times (\dfrac{a+4b}{25}+\dfrac{2a+3b}{25})\times \dfrac{h}{5}=55$ or $h=\dfrac{250\times {55}}{5(a+5b)}$ . Therefore $a=\dfrac{b}{2}, h=\dfrac{500}{b}$ and the required area is $\dfrac{1}{2}(a+b)h=\boxed {375}$