For a triangle XYZ, we use [XYZ] to denote its area.
Let ABCD be a square with side length 1. Points E and F lie on line BC and line CD, respectively, in such a way that angle EAF=45 degrees If [CEF]=1/9, what is the value of [AEF]?
Hello NerdyKid!
[ABCD] = \(1m^2\)
[CEF] = \(\frac{1}{9}m^2\)
\([CEF]=\frac{\overline{FC}^2}{2}=\frac{1}{9}m^2\\ \overline{FC}=\sqrt{\frac{2}{9}}m=\frac{\sqrt{2}}{3}m\\ \overline{DF}=1m-\overline{FC}=(1-\frac{\sqrt{2}}{3})m\)
\(2[ADF]=(1-\sqrt{\frac{2}{9}})m\cdot 1m=(1-\frac{\sqrt{2}}{3})m^2\)
\([AEF]=1m^2-[CEF]-2[ADF]\\ [AEF]=1m^2-\frac{1}{9}m^2-(1-\frac{ \sqrt{2}}{3})m^2 \\ [AEF]=1m^2-\frac{1}{9}m^2-1m^2+\frac{3\cdot \sqrt{2}}{9}m^2\\ [AEF]=\frac{3\cdot \sqrt{2}\ -1}{9} m^2 \\ \color{blue}[AEF]\approx0.3602934\ m^2\)
I have to recalculate that.
asinus
I calculated it. It is true.
!
