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# Quick Question

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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!

Oct 14, 2022

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Find the area of triangle ACE.

Hello Qube73!

$$\overline{AE}=\sqrt{22^2+10^2}=24.166\\ \angle CAE=\ atan \frac{10}{22}=24.444^\circ\\ \overline{CE}=\sqrt{\overline{AE}^2+30^2-2\cdot 24.166\cdot 30\cdot cos(\angle\ CAE)}\\ \overline{CE}=\sqrt{24.166^2+30^2-2\cdot 24.166\cdot 30\cdot cos\ 24.444^\circ}\\ \overline{CE}=12.806\ \color{Unnecessary !}$$

$$A_{ACE}=\frac{1}{2}\cdot 30\cdot \overline{AE}\cdot sin\ \angle CAE\\ A_{ACE}=\frac{1}{2}\cdot 30\cdot 24.166\cdot sin\ 24.444^\circ\\ \color{blue}A_{ACE}=150$$

!

Oct 14, 2022
edited by asinus  Oct 14, 2022
edited by asinus  Oct 14, 2022
#2
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That answer is wrong, it's actually 144.  But thanks for trying to help

--Qube73

Guest Oct 16, 2022
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Hello Qube73!

I checked my solution again and got 150. Please explain me your solution 144.

!

asinus  Oct 16, 2022
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Thanks Qube73 for acknowledging Asinus's solution.

I have done it independantly of asinus and got the same answer.

So maybe you would like to explain why you believe our answer is incorrect.

Oct 16, 2022
edited by Melody  Oct 16, 2022
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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.

Oct 16, 2022
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Thanks for the correction. It's all right.

!

asinus  Oct 16, 2022
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Thanks very much for clarifying Qube73  :)

Melody  Oct 16, 2022