+47 Hey y'all, this is my first time using this site, so I apologize if this isn't great.
So, I'm having some trouble with this problem,
"In triangle ABC, M is the midpoint of AB. Let D be the point on BC such that AD bisects angle BAC, and let the perpendicular bisector of AB intersect AD at E. If AB = 44, AC = 30, and ME = 10, then find the area of triangle ACE."
Does anyone know,
(A - How to do this problem?
(B - What the answer is?
Any assistance is greatly appreciated!
+14913 Find the area of triangle ACE.
Hello Qube73!
\(\overline{AE}=\sqrt{22^2+10^2}=24.166\\ \angle CAE=\ atan \frac{10}{22}=24.444^\circ\\ \overline{CE}=\sqrt{\overline{AE}^2+30^2-2\cdot 24.166\cdot 30\cdot cos(\angle\ CAE)}\\ \overline{CE}=\sqrt{24.166^2+30^2-2\cdot 24.166\cdot 30\cdot cos\ 24.444^\circ}\\ \overline{CE}=12.806\ \color{Unnecessary !}\)
\(A_{ACE}=\frac{1}{2}\cdot 30\cdot \overline{AE}\cdot sin\ \angle CAE\\ A_{ACE}=\frac{1}{2}\cdot 30\cdot 24.166\cdot sin\ 24.444^\circ\\ \color{blue}A_{ACE}=150\)
!
+47 Hey y'all, that wasn't me... that question was posted at almost 1 AM my time, I wasn't even awake at that point. Asinus' solution was correct, so whoever said that it was incorrect, was
(A - Not me.
(B - Doing a different question.
I'm sorry for any incovience this may have caused.
