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In right triangle ABC, $\angle CAB$ is a right angle. Point $M$ is the midpoint of $\overline{BC}$. What is the number of centimeters in the length of median $\overline{AM}$? Express your answer as a decimal to the nearest tenth

 

 

 Apr 26, 2020
 #1
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BTW when you see a right triangle a thing I always do is imagine it is inscribed in a circle with the hypotenuse as the diameter. That's just what I do, no need to do what I do. But is always nice to do what I do, because doing what I do makes me feel like people exist. I like when people exist, especially people who do what I do.

 

Anyways, draw a segment from \(A\) to to a point on \(BC\) in such it way that it is perpendicular to the hypotenuse. We call the point of intersection on the hypotenuse \(L\)

 

This means \(AL\perp{BC}\)

 - BC = 5 because it is a 3-4-5 triangle.

 

By AA similarity, we know that triangle \(ALC\) is similar to triangle \(ABC\). To find AL, we write the following proportion:

\(\frac{AL}{3}=\frac{4}{5}\)

 

Solving for AL: 

1. \(5AL=12\)

2. \(AL=\frac{12}{5}\)

 

Now here is how you solve the rest:

1. Consider point \(M\), that is the midpoint of \(BC\). Because it is the midpoint, we know that \(CM=\frac{5}{2}\)

2. Use the pythagorean theorem to find \(CL\)     hint:\(\sqrt{AC^2-AL^2}\)

3. Now that you found \(CL\), we can find \(LM\) by evaluating \(CM-CL\)

4. Since you know \(AL\) and \(LM\), you can easily find \(AM\) by pythagorean theorem.     hint: \(\sqrt{AL^2+LM^2}\)

 

Yay

 Apr 27, 2020
edited by AnExtremelyLongName  Apr 27, 2020

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