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

(→‎Solution 2)
(→‎Solution 2)
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~richiedelgado
 
~richiedelgado
  
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==Solution 2 ==
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==Solution 2==
 
<asy>
 
<asy>
 
pair A = (0,0);
 
pair A = (0,0);
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draw(rightanglemark(B,P,D));
 
draw(rightanglemark(B,P,D));
 
</asy>
 
</asy>
 +
 +
The diagram is the same as solution 1, just constrcuted a line at <math>BD</math>
 +
 +
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.
 +
 +
By knowing these statements, we get <math>\triangle ABD</math> ≅ <math>\triangle BDC</math>.
 +
 +
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).
 +
 +
We get the area of <math>\triangle ADB</math> as 10 because of the area of the triangle > <math>bh/2</math>.
 +
 +
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>.
  
 
== See Also ==
 
== See Also ==

Revision as of 01:32, 18 November 2022

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

[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]

(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, so $BP = 4$.

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

~richiedelgado

Solution 2

[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 the same as solution 1, just constrcuted a line at $BD$

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

By knowing these statements, we get $\triangle ABD$ ≅ $\triangle BDC$.

Knowing that $AP=3$ ; $AB=5$, then we would know that $BP=4$ because $\triangle APB$ is a 3-4-5 right triangle (as stated in solution 1).

We get the area of $\triangle ADB$ as 10 because of the area of the triangle > $bh/2$.

Since we know that $\triangle ABD$ ≅ $\triangle BDC$ and that $ABCD=\triangle ABD + \triangle BDC$. That means we double the area of $\triangle ADB$, $10 \cdot 2 = \boxed{\textbf{(D) }20}$.

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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