Difference between revisions of "2022 AMC 10B Problems/Problem 2"
(→Solution) |
Ehuang0531 (talk | contribs) (restored original not-to-scale figure in problem statement and moved to-scale figure to solution; ce) |
||
| Line 3: | Line 3: | ||
In rhombus <math>ABCD</math>, point <math>P</math> lies on segment <math>\overline{AD}</math> so that <math>\overline{BP}</math> <math>\perp</math> <math>\overline{AD}</math>, <math>AP = 3</math>, and <math>PD = 2</math>. What is the area of <math>ABCD</math>? (Note: The figure is not drawn to scale.) | In rhombus <math>ABCD</math>, point <math>P</math> lies on segment <math>\overline{AD}</math> so that <math>\overline{BP}</math> <math>\perp</math> <math>\overline{AD}</math>, <math>AP = 3</math>, and <math>PD = 2</math>. What is the area of <math>ABCD</math>? (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> | ||
| + | |||
| + | <math>\textbf{(A) }3\sqrt{5}\qquad\textbf{(B) }10\qquad\textbf{(C) }6\sqrt{5}\qquad\textbf{(D) }20\qquad\textbf{(E) }25</math> | ||
| + | |||
| + | ==Solution== | ||
| + | |||
<asy> | <asy> | ||
pair A = (0,0); | pair A = (0,0); | ||
| Line 20: | Line 37: | ||
</asy> | </asy> | ||
| − | + | (Figure redrawn to scale.) | |
| − | |||
| − | |||
| − | <math>AD = AP + PD = 3 + 2 =5</math> | + | <math>AD = AP + PD = 3 + 2 = 5.</math> |
| − | <math>ABCD</math> is a rhombus, so <math> | + | <math>ABCD</math> is a rhombus, so <math>AB = AP = 5</math>. |
<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 * 4 = \boxed{\textbf{(D) }20}</math>. | |
~richiedelgado | ~richiedelgado | ||
Revision as of 17:06, 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
