# Difference between revisions of "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