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

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

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.

@@ Line 14: / Line 14: @@
 ~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>
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Difference between revisions of "1956 AHSME Problems/Problem 27"

Latest revision as of 21:19, 12 February 2021

Problem 27

Solution

See Also