If $\Delta ABC$ is an isosceles triangle with $AB=AC=5, BC=6$ and has points $D, E \in AC$ such that $AD =1, DE=2, EC=2$, then the extent (or area) of $\Delta BDE = S$ computes to: $S = [\Delta ABC] - [\Delta ABD] - [\Delta BEC] = \frac{1}{2}(AB)(BC)sin \angle{B} - \frac{1}{2} (AB)(AD)sin \angle{A} - \frac{1}{2}(BC)(EC)sin \angle{C}$; or $S = \frac{1}{2}[(5)(6) \cdot (4/5) - (5)(1)(6/5)(4/5) - (6)(2)(4/5)]$; or $S = \frac{1}{2} \cdot \left(24 -\frac{24}{5}-\frac{48}{5}\right) = \frac{24}{5}$; or $15S =$$\fbox{72}$.

According to Heron's formula, the area of triangle ABC is 12. Because √(8 x 3 x 3 x 2)=√144=12. The length of line segment DE is 5−1−2=2, so S= 2/5 x 12= 24/5. Therefore, the answer is 24/5 x 15=72.