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

Revision as of 19:29, 19 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$",P,S); [/asy]$

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

Video Solution 1

https://youtu.be/Io_GhJ6Zr_U

~Education, the Study of Everything

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$", P, S); draw(A--B--C--D--cycle); draw(P--B); draw(rightanglemark(B,P,D)); [/asy]$

(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 (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 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 $\angle ABD$ $\cong$ $\angle BDC$ by the Alternate Interior Angles Theorem.

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

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 $\triangle ABD$ as 10 because of the area of the triangle -> $bh/2$ .

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

~ghfhgvghj10

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.

@@ Line 24: / Line 24: @@
 \textbf{(D) }20\qquad
 \textbf{(E) }25</math>
+==Video Solution 1==
+https://youtu.be/Io_GhJ6Zr_U
+~Education, the Study of Everything
 ==Solution 1==

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