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

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==Problem==
 
==Problem==
IDK Don't ask me.  
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A square with side length <math>3</math> is inscribed in an isosceles triangle with one side of the square along the base of the triangle. A square with side length <math>2</math> 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?
  
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(The official problem came with a daigram...)
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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>
  
 
==Solution==
 
==Solution==

Revision as of 10:58, 23 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?

(The official problem came with a daigram...)

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

Solution

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 (ProblemsAnswer KeyResources)
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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