Question:

In $\triangle ABC$, seg $AP$ is a median. If $BC = 18$, and $AB^2 + AC^2 = 260$, find $AP$.

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For any triangle, the median length can be found using the formula $AB^2 + AC^2 = 2(AP^2 + \tfrac{1}{4}BC^2)$.
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Solution and Explanation

Step 1: Recall the formula for the length of a median.
If $AP$ is a median to side $BC$ in $\triangle ABC$, then \[ AB^2 + AC^2 = 2(AP^2 + \tfrac{1}{4} BC^2) \] Step 2: Substitute the given values.
\[ 260 = 2(AP^2 + \tfrac{1}{4} \times 18^2) \] \[ 260 = 2(AP^2 + \tfrac{1}{4} \times 324) \] \[ 260 = 2(AP^2 + 81) \] Step 3: Simplify the equation.
\[ 260 = 2AP^2 + 162 \] \[ 260 - 162 = 2AP^2 \] \[ 98 = 2AP^2 \] \[ AP^2 = 49 \] Step 4: Find AP.
\[ AP = \sqrt{49} = 7 \] Step 5: Conclusion.
Therefore, the length of median $AP$ is 7 cm.
Correct Answer: $AP = 7$ cm
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