Divide an isosceles triangle

How to divide an isosceles triangle into two parts with equal contents perpendicular to the axis of symmetry (into a trapezoid and a triangle)?

Correct answer:

x =  1.4142

Step-by-step explanation:

$$\begin{gathered} h_2:a_2=h:a \\ {h}_2=h\cdot a_2\mathrm{/}a \\ {S}_1={\displaystyle \frac{h_2\cdot a_2}{2}}={\displaystyle \frac{a_2^2\cdot h}{2\cdot a}} \\ {S}_2={\displaystyle \frac{a+a_2}{2}}\cdot h(1-a_2\mathrm{/}a) \\ {S}_1=S_2 \\ {\displaystyle \frac{a_2^2\cdot h}{2\cdot a}}={\displaystyle \frac{a+a_2}{2}}\cdot h(1-a_2\mathrm{/}a) \\ {a}_2^2=(a+2)\cdot (a-a_2)=0 \\ {a}_2^2=a^2-a_2^2 \\ 2a_2^2=a^2 \\ \sqrt{2}{a}_2=a \\ {h}_2=h\cdot a_2\mathrm{/}a={\displaystyle \frac{a_2}{\sqrt{2}\cdot a_2}}=1\mathrm{/}\sqrt{2} \\ x=h:h_2 \\ x=\sqrt{2}=1.4142 \end{gathered}$$

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