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2022 AMC 10B Problems/Problem 2: Difference between revisions

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==Problem==
==Solution Made Complicated==


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.)
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>.


<asy>
Now, by spliting the rhombus into
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);
label("$B$", B, NW);
label("$C$",C,NE);
label("$D$",D,SE);
label("$P$",P,S);
</asy>


<math>\textbf{(A) }3\sqrt{5}\qquad\textbf{(B) }10\qquad\textbf{(C) }6\sqrt{5}\qquad\textbf{(D) }20\qquad\textbf{(E) }25</math>
==Solution 2==


==Solution==
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>, <imath>BP = 4</imath>.
<asy>
Therefore, <cmath>
pair A = (0,0);
[ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}}
label("$A$", A, SW);
</cmath>
pair B = (2.25,3);
~DUODUOLUO(First time using <imath>\LaTeX</imath>, may have some minor mistakes)
label("$B$", B, NW);
pair C = (6,3);
label("$C$", C, NE);
pair D = (3.75,0);
label("$D$", D, SE);
pair P = (2.25,0);
label("$P$", P, S);
draw(A--B--C--D--cycle);
draw(P--B);
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 = AP = 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
 
== 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}}

Latest revision as of 12:58, 10 November 2025

Solution Made Complicated

We find that $AP + DP = 5 = AD$. Because $ABCD$ is a rhombus, we get that AD = AB = 5. Notice that using Pythagorean Theorem, we have that $AB^2 = AP^2 + BP^2$, which simplifies to $25 = AP^2 + 9 ==> AP = \sqrt{16} = 4$.

Now, by spliting the rhombus into

Solution 2

Notice that $BC = AD = AP + PD = 3 + 2 = 5$. Now, because the question tells us $ABCD$ is a rhombus, $BC = AB = 5$. Thus, by Pythagorean Theorem on $APB$, $BP = 4$. Therefore, \[[ABCD] = AD \times BP = 5 \times 4 = \boxed{\text{D 20}}\] ~DUODUOLUO(First time using $\LaTeX$, may have some minor mistakes)

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