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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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| |
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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>
| | ==Solution 2== |
| import olympiad;
| |
| size(180);
| |
| real r = 3, s = 5, t = sqrt(r*r+s*s);
| |
| defaultpen(linewidth(0.6) + fontsize(10));
| |
| pair A = (0,0), B = (r,s), C = (r+t,s), D = (t,0), P = (r,0);
| |
| draw(A--B--C--D--A^^B--P^^rightanglemark(B,P,D));
| |
| 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);
| |
| </asy>
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| <math>\textbf{(A) }3\sqrt 5 \qquad | | Notice that <imath>BC = AD = AP + PD = 3 + 2 = 5</imath>. |
| \textbf{(B) }10 \qquad
| | 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>. |
| \textbf{(C) }6\sqrt 5 \qquad
| | Therefore, <cmath> |
| \textbf{(D) }20\qquad
| | [ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}} |
| \textbf{(E) }25</math>
| | </cmath> |
| | | ~DUODUOLUO(First time using <imath>\LaTeX</imath>, may have some minor mistakes) |
| ==Solution==
| |
| | |
| <asy>
| |
| pair A = (0,0);
| |
| label("$A$", A, SW);
| |
| 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));
| |
| </asy>
| |
| | |
| (Figure redrawn to scale.)
| |
| | |
| <math>AD = AP + PD = 3 + 2 = 5</math>.
| |
| | |
| <math>ABCD</math> is a rhombus, so <math>AB = AD = 5</math>. | |
| | |
| <math>\bigtriangleup APB</math> is a 3-4-5 right triangle, so <math>BP = 4</math>. | |
| | |
| The area of the rhombus <math>= bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}</math>.
| |
| | |
| ~richiedelgado | |
| | |
| ==Solution 2 The Area Of A Triangles==
| |
| <asy>
| |
| pair A = (0,0);
| |
| 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(D--B);
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| draw(B--P);
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| draw(rightanglemark(B,P,D));
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| </asy>
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| | |
| The diagram is the same as solution 1, just constrcuted a line at <math>BD</math>
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| | |
| When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. Which means <math>\angle ABD</math> ≅ <math>\angle BDC</math> by the Alternate Interior Angles Theorem.
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| | |
| By knowing these statements, we get <math>\triangle ABD</math> ≅ <math>\triangle BDC</math>.
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| | |
| Knowing that <math>AP=3</math> ; <math>AB=5</math>, then we would know that <math>BP=4</math> because <math>\triangle APB</math> is a 3-4-5 right triangle (as stated in solution 1).
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| | |
| We get the area of <math>\triangle ADB</math> as 10 because of the area of the triangle > <math>bh/2</math>.
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| | |
| Since we know that <math>\triangle ABD</math> ≅ <math>\triangle BDC</math> and that <math>ABCD=\triangle ABD + \triangle BDC</math>. That means we double the area of <math>\triangle ADB</math>, <math>10 \cdot 2 = \boxed{\textbf{(D) }20}</math>.
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| | |
| == See Also ==
| |
| | |
| {{AMC10 box|year=2022|ab=B|num-b=1|num-a=3}}
| |
| {{AMC12 box|year=2022|ab=B|num-b=1|num-a=3}}
| |
| {{MAA Notice}}
| |
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)
