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>. | ||
Line 7: | Line 12: | ||
<cmath>\boxed {C}</cmath> | <cmath>\boxed {C}</cmath> | ||
− | ~JustinLee2017 | + | ~JustinLee2017 |
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+ | ==See Also== | ||
+ | |||
+ | {{AHSME box|year=1956|num-b=26|num-a=28}} | ||
+ | |||
+ | {{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:
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
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 |
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