Question 10In triangle $ABC$, $AB=5$, $BC=8$ and $CA=11$. Point $D$ lies on $BC$ and $AD$ bisects $\angle BAC$. Find the area of triangle $ACD$.$\frac{4\sqrt{21}}5$$\frac{4\sqrt{21}}{11}$$\frac{11\sqrt{21}}4$$\frac{5\sqrt{21}}4$$\frac{11\sqrt5}4$