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Difference between revisions of "2021 Fall AMC 10B Problems/Problem 13"

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label("$C$",(1.5,3.3333333),SW);
 
label("$C$",(1.5,3.3333333),SW);
 
label("$D$",D,SE);
 
label("$D$",D,SE);
−
label("$H$",H,SE);
+
label("$E$",H,SE);
−
label("$J$",J,SE);
+
label("$F$",J,SE);
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label("$K$",K,N);
+
label("$G$",K,N);
  
  
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</asy>
 
</asy>
  
−
We see that $\bigtriangleup
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We see that <math>\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG</math> by AA similarity.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 17:52, 24 November 2021

Problem

A square with side length $3$ is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length $2$ 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]


$(\textbf{A})\: 19\frac14\qquad(\textbf{B}) \: 20\frac14\qquad(\textbf{C}) \: 21 \frac34\qquad(\textbf{D}) \: 22\frac12\qquad(\textbf{E}) \: 23\frac34$


Solution 1

Let's split the triangle down the middle and label it:

[asy] import olympiad; pair A,B,C,D,E,F,G,H,I,J,K; A = origin; B = (0.5,0); C=(2.5,0); D=(3,0); E = (0.5,2); F=(0.83333333333,2); G=(2.166666666667,2); H=(2.5,2); I=(0.83333333333,3.333333333333); J=(2.166666666667,3.3333333333); K=(1.5,6); draw(A--D--K--cycle); draw(B--E); draw(C--H); draw(F--I); draw(G--J); draw(I--J); draw(E--H); draw(K--(1.5,0)); label("$A$",(1.5,0),S); label("$B$",(1.5,2),SW); label("$C$",(1.5,3.3333333),SW); label("$D$",D,SE); label("$E$",H,SE); label("$F$",J,SE); label("$G$",K,N); [/asy]

We see that $\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG$ by AA similarity.

Solution 2

By similarity, the height is $3+\frac31\cdot2=9$ and the base is $\frac92\cdot1=4.5$. Thus the area is $\frac{9\cdot4.5}2=20.25=20\frac14$, or $\boxed{(\textbf{B})}$.

~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

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