Difference between revisions of "1956 AHSME Problems/Problem 27"

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== Problem 27==
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If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:
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<math>\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ \text{more than }6 </math>
 
==Solution==
 
==Solution==
 
Let the angle be <math>\theta</math> and the sides around it be <math>a</math> and <math>b</math>.  
 
Let the angle be <math>\theta</math> and the sides around it be <math>a</math> and <math>b</math>.  
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<cmath>\boxed {C}</cmath>
 
<cmath>\boxed {C}</cmath>
  
~JustinLee2017
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~JustinLee2017  
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==See Also==
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{{AHSME box|year=1956|num-b=26|num-a=28}}
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{{MAA Notice}}

Latest revision as of 21:19, 12 February 2021

Problem 27

If an angle of a triangle remains unchanged but each of its two including sides is doubled, then the area is multiplied by:

$\textbf{(A)}\ 2 \qquad\textbf{(B)}\ 3 \qquad\textbf{(C)}\ 4 \qquad\textbf{(D)}\ 6 \qquad\textbf{(E)}\ \text{more than }6$

Solution

Let the angle be $\theta$ and the sides around it be $a$ and $b$. The area of the triangle can be written as \[A =\frac{a \cdot b \cdot \sin(\theta)}{2}\] The doubled sides have length $2a$ and $2b$, while the angle is still $\theta$. Thus, the area is \[\frac{2a \cdot 2b \cdot \sin(\theta)}{2}\] \[\Rrightarrow  \frac{4ab \sin \theta}{2} = 4A\] \[\boxed {C}\]

~JustinLee2017

See Also

1956 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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 26 27 28 29 30
All AHSME Problems and Solutions


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