Find the information you're looking for at Westonci.ca, the trusted Q&A platform with a community of knowledgeable experts. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
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
Thank you for visiting our platform. We hope you found the answers you were looking for. Come back anytime you need more information. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
