$13$ Planar Surfaces, each of area $1\ m^{2}$ are such that their total area is $7\ m^{2}$. If no three of more of these surfaces have any region in common; Then, the overlap of some Two of These surfaces has an Area.

Assume are of overlap of any $2$ surfaces is less than $\dfrac{1}{13}m^2$

Let us ignore the unit of area in our notation.

We know that, $Area(A_{1}\cup A_{2}) = Area(A_{1}) + Area(A_{2}) - Area(A_{1}\cap A_{2})$

But, $Area (A_{1}) = Area (A_{2}) = 1.$ (as given in question)

Also, $Area(A_{1}\cap A_{2}) < \dfrac {1}{13}$ (as per our assumption)

Hence, $Area(A_{1}\cup A_{2}) > 1 + 1 - \dfrac {1}{13}$

$Area(A_{1}\cup A_{2}) > 2 - \dfrac {1}{13}$

Similarly,

$Area (A_{1}\cup A_{2}\cup A_{3}) = Area (A_{1}) + Area(A_{2}) + Area(A_{3})$

$- Area(A_{1}\cap A_{2})- Area (A_{1}\cap A_{3}) - Area(A_{2} \cap A_{3})$

$+ Area (A_{1}\cap A_{2}\cap A_{3})$

But $Area(A_{1}\cap A_{2}\cap A_{3}) = 0$ (As per the condition in Question)

Thus, $Area(A_{1}\cup A_{2}\cup A_{3}) > 1 + 1 + 1 - \dfrac {1}{13} - \dfrac {1}{13} - \dfrac {1}{13} + 0$.

i.e. $Area(A_{1} \cup A_{2} \cup A_{3}) > 3 - \dfrac {3}{13}$.

Similarly,

$Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) = Area(A_{1}) + Area (A_{2}) + Area(A_{3}) + Area(A_{4})$

$- Area(A_{1}\cap A_{2}) - Area(A_{1}\cap A_{3}) - Area (A_{1} \cap A_{4})$

$- Area(A_{2} \cap A_{3}) - Area(A_{2} \cap A_{4}) - Area (A_{3} \cap A_{4})$

$+ Area(A_{1}\cap A_{2}\cap A_{3})+ Area(A_{1}\cap A_{2}\cap A_{4})$

$+ Area (A_{2}\cap A_{3}\cap A_{4}) - Area (A_{1}\cap A_{2}\cap A_{3}\cap A_{4})$

But $Area(A_{1}\cap A_{2}\cap A_{3}\cap A_{4}) = 0$

$Area (A_{1} \cap A_{2}\cap A_{3}) = 0$, etc. (as per the condition in Question)

So, $Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) > 1 + 1 + 1 + 1 - 6\left (\dfrac {1}{13}\right ) + 3(0) - 0$.

i.e. $Area(A_{1}\cup A_{2}\cup A_{3}\cup A_{4}) > 4 - \dfrac {6}{13}$

In general,

$Area(A_{1}\cup A_{2}\cup A_{3}\cup ...\cup A_{n}) > n - \dfrac {^{n}C_{2}}{13}$

Put $n = 13$;

$Area(A_{1}\cup A_{2}\cup A_{3}\cup ....\cup A_{13}) > 13 - \dfrac {^{13}C_{2}}{13}$

That is $Area(A_{1}\cup A_{2}\cup A_{3}\cup ....\cup A_{13}) > 7$

Hence, our assumption was wrong.

So, Area of overlap of any $2$ surfaces is not less than $\dfrac {1}{13}$.