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# pls help

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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

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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