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| {{duplicate|[[2022 AMC 10B Problems/Problem 2|2022 AMC 10B #2]] and [[2022 AMC 12B Problems/Problem 2|2022 AMC 12B #2]]}}
| | ==Solution Made Complicated== |
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| ==Problem== | | We find that <imath>AP + DP = 5 = AD</imath>. |
| | 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{16} = 4 </imath>. |
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| In rhombus <math>ABCD</math>, point <math>P</math> lies on segment <math>\overline{AD}</math> so that <math>\overline{BP}</math> <math>\perp</math> <math>\overline{AD}</math>, <math>AP = 3</math>, and <math>PD = 2</math>. What is the area of <math>ABCD</math>? (Note: The figure is not drawn to scale.)
| | Now, by spliting the rhombus into |
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| <asy>
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| import olympiad;
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| size(180);
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| real r = 3, s = 5, t = sqrt(r*r+s*s);
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| defaultpen(linewidth(0.6) + fontsize(10));
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| pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0);
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| draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D));
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| label("$A$",A,SW);
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| label("$B$", B, NW);
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| label("$C$",C,NE);
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| label("$D$",D,SE);
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| label("$P$",P,S);
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| </asy>
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| <math>\textbf{(A) }3\sqrt 5 \qquad
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| \textbf{(B) }10 \qquad
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| \textbf{(C) }6\sqrt 5 \qquad
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| \textbf{(D) }20\qquad
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| \textbf{(E) }25</math>
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| ==Solution==
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| <asy>
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| pair A = (0,0);
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| label("$A$", A, SW);
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| pair B = (2.25,3);
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| label("$B$", B, NW);
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| pair C = (6,3);
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| label("$C$", C, NE);
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| pair D = (3.75,0);
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| label("$D$", D, SE);
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| pair P = (2.25,0);
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| label("$P$", P, S);
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| draw(A--B--C--D--cycle);
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| draw(P--B);
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| draw(rightanglemark(B,P,D));
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| </asy>
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| (Figure redrawn to scale.)
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| <math>AD = AP + PD = 3 + 2 = 5</math>.
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| <math>ABCD</math> is a rhombus, so <math>AB = AD = 5</math>.
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| <math>\bigtriangleup APB</math> is a 3-4-5 right triangle, so <math>BP = 4</math>.
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| The area of the rhombus <math>= bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}</math>.
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| ~richiedelgado
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| ==Solution 2== | | ==Solution 2== |
| Draw a line next to <math>BD</math>
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|
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| == See Also ==
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| {{AMC10 box|year=2022|ab=B|num-b=1|num-a=3}}
| | Notice that <imath>BC = AD = AP + PD = 3 + 2 = 5</imath>. |
| {{AMC12 box|year=2022|ab=B|num-b=1|num-a=3}}
| | 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>. |
| {{MAA Notice}} | | 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) |
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, ![\[[ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}}\]](//latex-new.aopstest.com/f/0/4/f043dfd0f57fde78d992447af9a1554fdcd1b0b7.png)
~DUODUOLUO(First time using 
, may have some minor mistakes)
.
Because 
is a rhombus, we get that AD = AB = 5.
Notice that using Pythagorean Theorem, we have that 
, which simplifies to 
.
.
Now, because the question tells us 
. 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 have some minor mistakes)
