1956 AHSME Problems/Problem 27

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

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1956 AHSME Problems/Problem 27

Solution