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: | ||
+ | |||
+ | <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 | + | ~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
[hide]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 The doubled sides have length and , while the angle is still . Thus, the area is
~JustinLee2017
Solution
Plugging into the quadratic formula, we get 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.