Difference between revisions of "1956 AHSME Problems/Problem 27"
(Added solution) |
Coolmath34 (talk | contribs) |
||
| Line 1: | Line 1: | ||
| + | == 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: | ||
| + | |||
| + | <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>. | ||
| Line 7: | Line 12: | ||
<cmath>\boxed {C}</cmath> | <cmath>\boxed {C}</cmath> | ||
| − | ~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== | ||
| + | |||
| + | {{AHSME box|year=1956|num-b=26|num-a=28}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 21:18, 12 February 2021
Contents
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:
Solution
Let the angle be 
and the sides around it be 
and 
.
The area of the triangle can be written as ![\[A =\frac{a \cdot b \cdot \sin(\theta)}{2}\]](http://latex.artofproblemsolving.com/f/4/3/f4351ef2c0dee5f30be832935220e90867f1ba01.png)
The doubled sides have length 
and 
, while the angle is still . Thus, the area is
![\[\frac{2a \cdot 2b \cdot \sin(\theta)}{2}\]](http://latex.artofproblemsolving.com/3/9/0/390685ef8590ce61cc3efaecc3584d241561f5eb.png)
![\[\Rrightarrow \frac{4ab \sin \theta}{2} = 4A\]](http://latex.artofproblemsolving.com/8/6/7/86703e392c386dfa6c7aa27865862857f191fc6b.png)
![\[\boxed {C}\]](http://latex.artofproblemsolving.com/9/d/1/9d1e11b199804f447fcd3a22d97c17ebbe9f963f.png)
~JustinLee2017
Solution
Plugging into the quadratic formula, we get
![\[x = \frac{2\sqrt{2} \pm \sqrt{8-4ac}}{2a}.\]](http://latex.artofproblemsolving.com/c/8/d/c8d83dfb1fbba5bc64524dbfbf0aa38fe0a5325d.png)
The discriminant is equal to 0, so this simplifies to 
Because we are given that is real, 
is always rational and the answer is 
See Also
| 1956 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
