2022 AMC 10B Problems/Problem 2: Difference between revisions
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Now, by spliting the rhombus into | Now, by spliting the rhombus into | ||
==Solution 2== | |||
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. | |||
Therefore <imath>[ABCD] = AD x BP = 5 x 4 &= \boxed{\textbf{(D)}\ {20}. | |||
\end{align*}</imath>$ | |||
Revision as of 23:31, 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 $[ABCD] = AD x BP = 5 x 4 &= \boxed{\textbf{(D)}\ {20}.
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)$
