\[x\]

.We must find an equation in terms of

\[x\]

for the length \[l\]

of the cutting line. When \[x=0, \: l=10\]

and when \[x=4, \: l=4\]

.From these two,

\[l=10-1.5x\]

The area of the part above the cutting line is

\[\frac{1}{2}(4+10-1.5x)(4-x)\]

The area of the part below the cutting line is

\[\frac{1}{2}(10+10-1.5x)x\]

Equating these gives

\[\frac{1}{2}(4+10-1.5x)(4-x)=\frac{1}{2}(10+10-1.5x)x\]

Expanding this and simplifying gives

\[3x^2-40x+56=0\]

Solving this equation gives

\[x= \frac{40 \pm \sqrt{928}}{6}\]

Obviously x must be less than the height of the trapezium so

\[x= \frac{40 - \sqrt{928}}{6}= x= \frac{20 - 2\sqrt{58}}{3}\]

since the other possibility gives \[x \gt 4\]

.