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2022 AMC 10B Problems/Problem 2: Difference between revisions

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


Now, by spliting the rhombus into
Now, by spliting the rhombus into
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[ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}}
[ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}}
</cmath>
</cmath>
~DUODUOLUO(First time using <imath>\LaTeX</imath>, may hav esome minor mistakes)
~DUODUOLUO(First time using <imath>\LaTeX</imath>, may have some minor mistakes)

Latest revision as of 12:58, 10 November 2025

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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