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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 <imath>ABCD</imath>, point <imath>P</imath> lies on segment <imath>\overline{AD}</imath> so that <imath>\overline{BP}</imath> <imath>\perp</imath> <imath>\overline{AD}</imath>, <imath>AP = 3</imath>, and <imath>PD = 2</imath>. What is the area of <imath>ABCD</imath>? (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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| <imath>\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</imath>
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| ==Solution 1==
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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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| <cmath>\textbf{Figure redrawn to scale.}</cmath>
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| <imath>AD = AP + PD = 3 + 2 = 5</imath>.
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| <imath>ABCD</imath> is a rhombus, so <imath>AB = AD = 5</imath>.
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| <imath>\bigtriangleup APB</imath> is a <imath>3-4-5</imath> right triangle, hence <imath>BP = 4</imath>.
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| The area of the rhombus is base times height: <imath>bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}</imath>.
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| ~richiedelgado
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| ==Solution 2 (The Area Of A Triangle)==
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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(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 from as Solution 1, but a line is constructed at <imath>BD</imath>.
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| When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that <imath>\angle ABD \cong \angle BDC</imath>, by the Alternate Interior Angles Theorem.
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| By SAS Congruence, we get <imath>\triangle ABD \cong \triangle BDC</imath>.
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| Since <imath>AP=3</imath> and <imath>AB=5</imath>, we know that <imath>BP=4</imath> because <imath>\triangle APB</imath> is a 3-4-5 right triangle, as stated in Solution 1.
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| The area of <imath>\triangle ABD</imath> would be <imath>10</imath>, since the area of the triangle is <imath>\frac{bh}{2}</imath>.
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| Since we know that <imath>\triangle ABD \cong \triangle BDC</imath> and that <imath>ABCD=\triangle ABD + \triangle BDC</imath>, so we can double the area of <imath>\triangle ADB</imath> to get <imath>10 \cdot 2 = \boxed{\textbf{(D) }20}</imath>.
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| ~ghfhgvghj10, minor edits by MinecraftPlayer404
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| ==Video Solution 1==
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| https://youtu.be/Io_GhJ6Zr_U
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| ~Education, the Study of Everything
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| ==Video Solution(1-16)==
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| https://youtu.be/SCwQ9jUfr0g
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| ~~Hayabusa1
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| ==Video Solution by Interstigation==
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| https://youtu.be/_KNR0JV5rdI?t=97
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| == See Also ==
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| {{AMC10 box|year=2022|ab=B|num-b=1|num-a=3}}
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| {{AMC12 box|year=2022|ab=B|num-b=1|num-a=3}}
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| {{MAA Notice}}
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.
Because 
is a rhombus, we get that AD = AB = 5.
Notice that using Pythagorean Theorem, we have that 
, which simplifies to 
.
