2021 Fall AMC 10B Problems/Problem 13: Difference between revisions
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<math>(\textbf{A})\: 19\frac14\qquad(\textbf{B}) \: 20\frac14\qquad(\textbf{C}) \: 21 \frac34\qquad(\textbf{D}) \: 22\frac12\qquad(\textbf{E}) \: 23\frac34</math> | <math>(\textbf{A})\: 19\frac14\qquad(\textbf{B}) \: 20\frac14\qquad(\textbf{C}) \: 21 \frac34\qquad(\textbf{D}) \: 22\frac12\qquad(\textbf{E}) \: 23\frac34</math> | ||
==Solution== | |||
==Solution 1== | |||
Let's split the triangle down the middle: | |||
<asy> | |||
import olympiad; | |||
pair A,B,C,D,E,F,G,H,I,J,K; | |||
A = origin; B = (0.25,0); C=(1.25,0); D=(1.5,0); E = (0.25,1); F=(0.4166666667,1); G=(1.08333333333,1); H=(1.25,1); I=(0.4166666667,1.66666666667); J=(1.08333333333,1.666666666667); K=(0.75,3); | |||
draw(A--D--K--cycle); | |||
draw(B--E); | |||
draw(C--H); | |||
draw(F--I); | |||
draw(G--J); | |||
draw(I--J); | |||
draw(E--H); | |||
draw(I--(0.75,0)); | |||
</asy> | |||
==Solution 2== | |||
By similarity, the height is <math>3+\frac31\cdot2=9</math> and the base is <math>\frac92\cdot1=4.5</math>. | By similarity, the height is <math>3+\frac31\cdot2=9</math> and the base is <math>\frac92\cdot1=4.5</math>. | ||
Thus the area is <math>\frac{9\cdot4.5}2=20.25=20\frac14</math>, or <math>\boxed{(\textbf{B})}</math>. | Thus the area is <math>\frac{9\cdot4.5}2=20.25=20\frac14</math>, or <math>\boxed{(\textbf{B})}</math>. | ||
Revision as of 15:36, 24 November 2021
Problem
A square with side length 
is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length 
has two vertices on the other square and the other two on sides of the triangle, as shown. What is the area of the triangle?
![[asy] import olympiad; pair A,B,C,D,E,F,G,H,I,J,K; A = origin; B = (0.25,0); C=(1.25,0); D=(1.5,0); E = (0.25,1); F=(0.4166666667,1); G=(1.08333333333,1); H=(1.25,1); I=(0.4166666667,1.66666666667); J=(1.08333333333,1.666666666667); K=(0.75,3); draw(A--D--K--cycle); draw(B--E); draw(C--H); draw(F--I); draw(G--J); draw(I--J); draw(E--H); [/asy]](http://latex.artofproblemsolving.com/2/0/c/20ced60f6d642645a5e5713fa7723356bdecff14.png)
Solution 1
Let's split the triangle down the middle:
![[asy] import olympiad; pair A,B,C,D,E,F,G,H,I,J,K; A = origin; B = (0.25,0); C=(1.25,0); D=(1.5,0); E = (0.25,1); F=(0.4166666667,1); G=(1.08333333333,1); H=(1.25,1); I=(0.4166666667,1.66666666667); J=(1.08333333333,1.666666666667); K=(0.75,3); draw(A--D--K--cycle); draw(B--E); draw(C--H); draw(F--I); draw(G--J); draw(I--J); draw(E--H); draw(I--(0.75,0)); [/asy]](http://latex.artofproblemsolving.com/c/4/1/c41c08f9121fd5857e8dd7b7cc2182a751df4e5d.png)
Solution 2
By similarity, the height is 
and the base is 
.
Thus the area is 
, or 
.
~Hefei417, or 陆畅 Sunny from China
See Also
| 2021 Fall AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America. 
