\(\displaystyle \text{Let the area of the triangle ABC be }\; \Delta,\\ \text{then }\; \frac{1}{2}AB.AC.\sin A = \Delta.\\ \text{So, since }\;AF=5AB/6,\; \text{and}\;AE=AC/4,\\ \text{the area of the triangle AFE=}\;\frac{1}{2}.\frac{5}{6}AB.\frac{1}{4}AC. \sin A = \frac{5}{24}\Delta.\)
Over to you.
