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Sagot :
A triangle is made of 3 parallel lines, in which all 3 corners have 3 faces (angles), that calls the vertex, in which two factions of a triangle get together, and one of its 3 sides can be the basis of a triangle, even so, generally the lower one.
Given:
[tex]\[ABCD\] = 1 m^2\\\\[/tex]
[tex][CEF ] = \frac{1}{9} m^2\\\\[/tex]
Find:
[tex]\[AEF\] =?[/tex]
Solution:
[tex][CEF] = \frac{\bar{FC}^2}{2} = \frac{1}{9} m^2\\\\\bar{FC}=\sqrt{\frac{2}{9}m}=\frac{\sqrt{2}}{3}m\\\\\bar{DF}= 1m -\bar{FC}=(1- \frac{\sqrt{2}}{3})m\\\\[/tex]
[tex]2[ ADF]= (1- \sqrt{\frac{2}{9}}) m \cdot 1 m =(1 -\frac{\sqrt{2}}{3})m^2 \\\\[/tex]
[tex][ AEF ]= 1 m^2- [CEF]-2[ ADF]\\\\[/tex]
[tex]= 1m^2- \frac{1}{9} m^2 -(1 -\frac{\sqrt{2}}{3})m^2\\\\= 1m^2- \frac{1}{9} m^2 -1m^2 +\frac{3\sqrt{2}}{9}m^2\\\\= \frac{ 3\sqrt{2} -1}{9}m^2\approx 0.3602934\ m^2[/tex]
So, the final answer is "[tex]\bold{0.3602934\ m^2}[/tex]".
Learn more:
brainly.com/question/10468052
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