# 1956 AHSME Problems/Problem 27

## 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

## Solution

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

## See Also

 1956 AHSME (Problems • Answer Key • Resources) Preceded byProblem 26 Followed byProblem 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.

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