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Difference between revisions of "2022 AMC 10B Problems/Problem 2"

(→‎Solution 2 (The Area Of A Triangle))
 
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</asy>
 
</asy>
  
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(Figure redrawn to scale.)
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<cmath>\textbf{Figure redrawn to scale.}</cmath>
  
 
<math>AD = AP + PD = 3 + 2 = 5</math>.
 
<math>AD = AP + PD = 3 + 2 = 5</math>.
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<math>ABCD</math> is a rhombus, so <math>AB = AD = 5</math>.
 
<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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<math>\bigtriangleup APB</math> is a <math>3-4-5</math> right triangle, hence <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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The area of the rhombus is base times height: <math>bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}</math>.
  
 
~richiedelgado
 
~richiedelgado
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</asy>
 
</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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The diagram is from as Solution 1, but a line is constructed 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>\cong</math> <math>\angle BDC</math> by the Alternate Interior Angles Theorem.  
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When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that <math>\angle ABD \cong \angle BDC</math>, by the Alternate Interior Angles Theorem.
  
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By SAS Congruence, we get <math>\triangle ABD</math> <math>\cong</math> <math>\triangle BDC</math>.
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By SAS Congruence, we get <math>\triangle ABD \cong \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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Since <math>AP=3</math> and <math>AB=5</math>, we 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 ABD</math> as 10 because of the area of the triangle -> <math>bh/2</math>.
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The area of <math>\triangle ABD</math> would be <math>10</math>, since the area of the triangle is <math>\frac{bh}{2}</math>.
  
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Since we know that <math>\triangle ABD</math> <math>\cong</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>.
+
Since we know that <math>\triangle ABD \cong \triangle BDC</math> and that <math>ABCD=\triangle ABD + \triangle BDC</math>, so we can double the area of <math>\triangle ADB</math> to get <math>10 \cdot 2 = \boxed{\textbf{(D) }20}</math>.
  
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~ghfhgvghj10
+
~ghfhgvghj10, minor edits by MinecraftPlayer404
 +
 
 +
==Video Solution 1==
 +
https://youtu.be/Io_GhJ6Zr_U
 +
 
 +
~Education, the Study of Everything
 +
 
 +
==Video Solution(1-16)==
 +
https://youtu.be/SCwQ9jUfr0g
 +
 
 +
~~Hayabusa1
 +
==Video Solution by Interstigation==
 +
https://youtu.be/_KNR0JV5rdI?t=97
  
 
== See Also ==
 
== See Also ==

Latest revision as of 16:14, 29 June 2023

The following problem is from both the 2022 AMC 10B #2 and 2022 AMC 12B #2, so both problems redirect to this page.

Problem

In rhombus $ABCD$, point $P$ lies on segment $\overline{AD}$ so that $\overline{BP}$ $\perp$ $\overline{AD}$, $AP = 3$, and $PD = 2$. What is the area of $ABCD$? (Note: The figure is not drawn to scale.)

[asy] 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]

$\textbf{(A) }3\sqrt 5 \qquad \textbf{(B) }10 \qquad \textbf{(C) }6\sqrt 5 \qquad \textbf{(D) }20\qquad \textbf{(E) }25$

Solution 1

[asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); 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]

\[\textbf{Figure redrawn to scale.}\]

$AD = AP + PD = 3 + 2 = 5$.

$ABCD$ is a rhombus, so $AB = AD = 5$.

$\bigtriangleup APB$ is a $3-4-5$ right triangle, hence $BP = 4$.

The area of the rhombus is base times height: $bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}$.

~richiedelgado

Solution 2 (The Area Of A Triangle)

[asy] pair A = (0,0); label("$A$", A, SW); pair B = (2.25,3); 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(D--B); draw(B--P); draw(rightanglemark(B,P,D)); [/asy]

The diagram is from as Solution 1, but a line is constructed at $BD$.

When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that $\angle ABD \cong \angle BDC$, by the Alternate Interior Angles Theorem.

By SAS Congruence, we get $\triangle ABD \cong \triangle BDC$.

Since $AP=3$ and $AB=5$, we know that $BP=4$ because $\triangle APB$ is a 3-4-5 right triangle, as stated in Solution 1.

The area of $\triangle ABD$ would be $10$, since the area of the triangle is $\frac{bh}{2}$.

Since we know that $\triangle ABD \cong \triangle BDC$ and that $ABCD=\triangle ABD + \triangle BDC$, so we can double the area of $\triangle ADB$ to get $10 \cdot 2 = \boxed{\textbf{(D) }20}$.

~ghfhgvghj10, minor edits by MinecraftPlayer404

Video Solution 1

https://youtu.be/Io_GhJ6Zr_U

~Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

Video Solution by Interstigation

https://youtu.be/_KNR0JV5rdI?t=97

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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