28 Three sides $O A B, O A C$ and $O B C$ of a tetrahedron $O A B C$ are right-angled triangles, i.e. $\angle A O B=\angle A O C=\angle B O C=90^{\circ}$. Given that $O A=7, O B=2$ and $O C=6$, find the value of
(Area of $\triangle O A B)^{2}+(\text { Area of } \triangle O A C)^{2}+(\text { Area of } \triangle O B C)^{2}+(\text { Area of } \triangle A B C)^{2}$.
$\begin{array}{l}28 \text { Answer: (1052) } \\ \text { Note that }(\text { Area of } \triangle O A B)^{2}=(\text { Area of } \triangle O A C)^{2}+(\text { Area of } \triangle O B C)^{2}+(\text { Area of } \triangle A B C)^{2} \text {. } \\ \text { So (Area of } \triangle O A B)^{2}+(\text { Area of } \triangle O A C)^{2}+(\text { Area of } \triangle O B C)^{2}+(\text { Area of } \triangle A B C)^{2} \\ =2 \times\left\{(\text { Area of } \triangle O A C)^{2}+(\text { Area of } \triangle O B C)^{2}+(\text { Area of } \triangle A B C)^{2}\right\} \\ =2 \times\left((1 / 2 \times 7 \times 6)^{2}+(1 / 2 \times 7 \times 2)^{2}+(1 / 2 \times 2 \times 6)^{2}\right\}=1052 \text {. }\end{array}$