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

 
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~JustinLee2017  
 
~JustinLee2017  
  
== Solution ==
 
Plugging into the quadratic formula, we get
 
<cmath>x = \frac{2\sqrt{2} \pm \sqrt{8-4ac}}{2a}.</cmath>
 
The discriminant is equal to 0, so this simplifies to <math>x = \frac{2\sqrt{2}}{2a}=\frac{\sqrt{2}}{a}.</math> Because we are given that <math>a</math> is real, <math>x</math> is always rational and the answer is <math>\boxed{\textbf{(B)}}.</math>
 
 
==See Also==
 
==See Also==
  

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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