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 | 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 | ||
| Line 10: | Line 10: | ||
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 < | Therefore, <cmath> | ||
\ | [ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}} | ||
</cmath> | |||
~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
.
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,
~DUODUOLUO(First time using
, may have some minor mistakes)