Divide an isosceles triangle

How can an isosceles triangle be divided into two parts with equal areas perpendicular to the axis of symmetry (into a trapezoid and a triangle)?

Correct answer:

x =  1.4142

Step-by-step explanation:

$$\begin{gathered} h_2:a_2=h:a \\ {h}_2=h\cdot a_2/a \\ {S}_1=\frac{h_2\cdot a_2}{2}=\frac{a_2^2\cdot h}{2\cdot a} \\ {S}_2=\frac{a+a_2}{2}\cdot h(1-a_2/a) \\ {S}_1=S_2 \\ \frac{a_2^2\cdot h}{2\cdot a}=\frac{a+a_2}{2}\cdot h(1-a_2/a) \\ {a}_2^2=(a+2)\cdot (a-a_2) \\ {a}_2^2=a^2-a_2^2 \\ 2a_2^2=a^2 \\ \sqrt{2}{a}_2=a \\ {h}_2=h\cdot a_2/a=\frac{a_2}{\sqrt{2}\cdot a_2}=1/\sqrt{2} \\ x=h:h_2 \\ x=\sqrt{2}=1.4142 \end{gathered}$$

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