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 < | Therefore <cmath>[ABCD] = AD \times BP = 5 \times 4 = \boxed{\textbf{(D)}\ 20} | ||
\end{align*}</ | \end{align*}</cmath> | ||
Revision as of 23:32, 9 November 2025
Solution Made Complicated
We find that 
.
Because 
is a rhombus, we get that AD = AB = 5.
Notice that using Pythagorean Theorem, we have that 
, which simplifies to 
.
Now, by spliting the rhombus into
Solution 2
Notice that 
.
Now, because the question tells us is a rhombus, 
. Thus, by Pythagorean Theorem on 
, BP = 4.
Therefore
\[[ABCD] = AD \times BP = 5 \times 4 = \boxed{\textbf{(D)}\ 20}
\end{align*}\] (Error compiling LaTeX. Unknown error_msg)
