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 $\boxed{\textbf{(B)}}.$

1956 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 26	Followed by Problem 28
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1956 AHSME Problems/Problem 27

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Problem 27

Solution

Solution

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