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

(→‎Solution 1)
(→‎Solution 1)
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==Solution 1==
 
==Solution 1==
  
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Let's split the triangle down the middle:
+
Let's split the triangle down the middle and label it:
  
 
<asy>
 
<asy>
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draw(E--H);
 
draw(E--H);
 
draw(K--(0.75,0));
 
draw(K--(0.75,0));
 +
label("$A$",A,W);
 +
label("$B$",B,S);
 +
label("$C$",C,S);
 +
label("$D$",D,E);
 +
label("$E$",E,W);
 +
label("$F$",F,S);
 +
label("$G$",G,S);
 +
label("$H$",H,E);
 +
label("$I$",I,W);
 +
label("$J$",J,E);
 +
label("$K$",K,N);
 +
  
  

Revision as of 15:38, 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 = (0.25,0); 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(K--(0.75,0)); label("$A$",A,W); label("$B$",B,S); label("$C$",C,S); label("$D$",D,E); label("$E$",E,W); label("$F$",F,S); label("$G$",G,S); label("$H$",H,E); label("$I$",I,W); label("$J$",J,E); label("$K$",K,N); [/asy]

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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