# What is the area of a triangle with sides of length 4, 6, and 9?

Let a = 4, b = 6, and c = 9. $A = {\cos}^{-} 1 \left(\frac{{6}^{2} + {9}^{2} - {4}^{2}}{2 \cdot 6 \cdot 9}\right) = 20.74 \ldots$
$A r e a = \frac{1}{2} b c \sin A = \frac{1}{2} \left(6\right) \left(9\right) \sin 20.74 \ldots = 9.56 u n i {t}^{2}$