2022 AMC 10B Problems/Problem 2: Difference between revisions
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==Problem== | ==Problem== | ||
In rhombus < | 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.) | ||
<asy> | <asy> | ||
| Line 19: | Line 19: | ||
</asy> | </asy> | ||
< | <imath>\textbf{(A) }3\sqrt 5 \qquad | ||
\textbf{(B) }10 \qquad | \textbf{(B) }10 \qquad | ||
\textbf{(C) }6\sqrt 5 \qquad | \textbf{(C) }6\sqrt 5 \qquad | ||
\textbf{(D) }20\qquad | \textbf{(D) }20\qquad | ||
\textbf{(E) }25</ | \textbf{(E) }25</imath> | ||
==Solution 1== | ==Solution 1== | ||
Revision as of 21:15, 9 November 2025
- 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 
, point 
lies on segment 
so that 
, 
, and 
. What is the area of ? (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]](http://latex.artofproblemsolving.com/1/5/0/150aeac95fbc38e4ea6602817ade871e681d99e3.png)
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]](http://latex.artofproblemsolving.com/0/f/7/0f78edcce7922133d86e208ab951ade856a91a0c.png)
![\[\textbf{Figure redrawn to scale.}\]](http://latex.artofproblemsolving.com/c/0/a/c0a02f10384763e46782543950ad10b71b449c73.png)
.
is a rhombus, so 
.
is a 
right triangle, hence 
.
The area of the rhombus is base times height: 
.
~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]](http://latex.artofproblemsolving.com/2/0/1/201486067897fc30746806b9e954f1ea20f6483c.png)
The diagram is from as Solution 1, but a line is constructed at 
.
When it comes to the sides of a rhombus, their opposite sides are congruent and parallel. This means that 
, by the Alternate Interior Angles Theorem.
By SAS Congruence, we get 
.
Since 
and 
, we know that 
because 
is a 3-4-5 right triangle, as stated in Solution 1.
The area of 
would be 
, since the area of the triangle is 
.
Since we know that and that 
, so we can double the area of 
to get 
.
~ghfhgvghj10, minor edits by MinecraftPlayer404
Video Solution 1
~Education, the Study of Everything
Video Solution(1-16)
~~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 | |
These problems are copyrighted © by the Mathematical Association of America. 
