Answer:
[tex]S_1+S_2+2\sqrt{S_1S_2}[/tex]
Step-by-step explanation:
Consider trapezoid ABCD. Let triangle BEC be of area [tex]S_1[/tex] and triangle AED be of area [tex]S_2.[/tex]
Triangles BEC and AED are similar with the coefficient of similarity [tex]\dfrac{A_{\triangle BEC}}{A_{\triangle AED}}=\sqrt{\dfrac{S_1}{S_2}}.[/tex]
Consider triangles ABC and ACD. The area of these triangles are
[tex]A_{\triangle ABC}=\dfrac{1}{2}\cdot BC\cdot H,\\ \\A_{\triangle ACD}=\dfrac{1}{2}\cdot AD\cdot H.[/tex]
Note that
[tex]H=h_{BC}+h_{AD},[/tex] where [tex]h_{BC}[/tex] is the height of the triangle BEC, [tex]h_{AD}[/tex] is the height of the triangle AED and
[tex]\dfrac{h_{BC}}{h_{AD}}=\sqrt{\dfrac{S_1}{S_2}}.[/tex]
Thus,
[tex]h_{BC}=\sqrt{\dfrac{S_1}{S_2}}\cdot h_{AD}[/tex] and
[tex]H=\left(1+\sqrt{\dfrac{S_1}{S_2}}\right)h_{AD}[/tex] or
[tex]H=\left(1+\sqrt{\dfrac{S_2}{S_1}}\right)h_{BC}.[/tex]
Then
[tex]A_{\triangle ACD}=\dfrac{1}{2}\cdot AD\cdot H=\dfrac{1}{2}\cdot AD\cdot\left(1+\sqrt{\dfrac{S_1}{S_2}}\right)h_{AD}=\left(1+\sqrt{\dfrac{S_1}{S_2}}\right)S_2,[/tex]
[tex]A_{\triangle ABC}=\dfrac{1}{2}\cdot BC\cdot H=\dfrac{1}{2}\cdot BC\cdot\left(1+\sqrt{\dfrac{S_2}{S_1}}\right)h_{BC}=\left(1+\sqrt{\dfrac{S_2}{S_1}}\right)S_1.[/tex]
Thus, the area of trapezoid ABCD is
[tex]A_{ABCD}=A_{\triangle ABC}+A_{\triangle ACD}=\left(1+\sqrt{\dfrac{S_1}{S_2}}\right)S_2+\left(1+\sqrt{\dfrac{S_2}{S_1}}\right)S_1=S_2+\sqrt{S_1S_2}+S_1+\sqrt{S_1S_2}=S_1+S_2+2\sqrt{S_1S_2}.[/tex]
