2022 AMC 10B Problems/Problem 2: Difference between revisions
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(Figure redrawn to scale.) | (Figure redrawn to scale.) | ||
<math>AD = AP + PD = 3 + 2 = 5 | <math>AD = AP + PD = 3 + 2 = 5</math>. | ||
<math>ABCD</math> is a rhombus, so <math>AB = AP = 5</math>. | <math>ABCD</math> is a rhombus, so <math>AB = AP = 5</math>. | ||
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<math>\bigtriangleup APB</math> is a 3-4-5 right triangle, so <math>BP = 4</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 | The area of the rhombus <math>= bh = (AD)(BP) = 5 \cdot 4 = \boxed{\textbf{(D) }20}</math>. | ||
~richiedelgado | ~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}} | |||
Revision as of 17:09, 17 November 2022
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
![[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)
(Figure redrawn to scale.)
.
is a rhombus, so 
.
is a 3-4-5 right triangle, so 
.
The area of the rhombus 
.
~richiedelgado
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. 
