Art of Problem Solving

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

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Notice that <imath>BC = AD = AP + PD = 3 + 2 = 5</imath>.
Notice that <imath>BC = AD = AP + PD = 3 + 2 = 5</imath>.
Now, because the question tells us <imath>ABCD</imath> is a rhombus, <imath>BC = AB = 5</imath>. Thus, by Pythagorean Theorem on <imath>APB</imath>, BP = 4.
Now, because the question tells us <imath>ABCD</imath> is a rhombus, <imath>BC = AB = 5</imath>. Thus, by Pythagorean Theorem on <imath>APB</imath>, BP = 4.
Therefore <cmath>[ABCD] = AD \times BP = 5 \times 4 = \boxed{\textbf{(D)}\ 20}
<cmath>
\end{align*}</cmath>
[ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}}
</cmath>
\end{align*}<imath></imath>

Revision as of 23:34, 9 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. \[[ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}}\] \end{align*}$$ (Error compiling LaTeX. Unknown error_msg)