1956 AHSME Problems/Problem 27

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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 \[\frac{2a \cdot 2b \cdot \sin(\theta)}{2}\] \[\Rrightarrow  \frac{4ab \sin \theta}{2} = 4A\] \[\boxed {C}\]

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