Let the number of pennies that Betty has be $ n.$ If after adding two more pennies, Betty can only arrange the pennies in a straight line, then $ n+2 $ must be prime. Since all primes greater than $ 2 $ are odd, $ n $ must be odd as well. Because she can arrange $ n $ pennies in three different ways, $ n $ must have four factors other than $ 1 $ and itself. Hence we examine the odd numbers between $ 40 $ and $ 50$. $ \bullet $ $ 41 $ factors as $ 1 \cdot 41 $ only. $ \bullet $ $ 43 $ factors as $ 1 \cdot 43 $ only. $ \bullet $ $ 45 $ factors as $ 1 \cdot 45 $ or $ 3 \cdot 15 $ or $ 5 \cdot 9 $. $ \bullet $ $ 47 $ factors as $ 1 \cdot 47 $ only. $ \bullet $ $ 49 $ factors as $ 1 \cdot 49 $ or $ 7 \cdot 7 $. The only number with four factors other than $ 1 $ or itself is $ 45$. If we add $ 2 $ to $ 45, $ we get $ 47, $ which is indeed prime. Hence, Betty has $ \boxed{45} $ pennies.