2022 AMC 10B Problems/Problem 2

Solution Made Complicated

We find that $AP + DP = 5 = AD$ . Because $ABCD$ is a rhombus, we get that AD = AB = 5. Notice that using Pythagorean Theorem, we have that $AB^2 = AP^2 + BP^2$ , which simplifies to $25 = AP^2 + 9 ==> AP = sqrt(16) = 4$ .

Now, by spliting the rhombus into

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2022 AMC 10B Problems/Problem 2

Solution Made Complicated