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

Solution 2

Notice that $BC = AD = AP + PD = 3 + 2 = 5$ . Now, because the question tells us $ABCD$ is a rhombus, $BC = AB = 5$ . Thus, by Pythagorean Theorem on $APB$ , $BP = 4$ . Therefore, $[ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}}$ ~DUODUOLUO(First time using $\LaTeX$ , may have some minor mistakes)

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2022 AMC 10B Problems/Problem 2

Solution Made Complicated

Solution 2