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>, <imath>BP = 4</imath>. | ||
Therefore, <cmath> | Therefore, <cmath> | ||
[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) | |||
Revision as of 23:36, 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 
, 
.
Therefore, ![\[[ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}}\]](http://latex-new.aopstest.com/f/0/4/f043dfd0f57fde78d992447af9a1554fdcd1b0b7.png)
~DUODUOLUO(First time using 
, may hav esome minor mistakes)
