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Standard XII

Mathematics

Union

13 Planar Sur...

Question

$13$ 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.

Not less than

\frac{7}{13} m^{2}

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Not less than

\frac{2}{13} m^{2}

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Not less than

\frac{1}{13} m^{2}

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Not Less than

\frac{1}{17} m^{2}

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Solution

The correct options are
C Not Less than $\frac{1}{17} m^{2}$
D Not less than $\frac{1}{13} m^{2}$
Assume are of overlap of any $2$ surfaces is less than $\frac{1}{13} m^{2}$
Let us ignore the unit of area in our notation.

We know that, $A r e a (A_{1} \cup A_{2}) = A r e a (A_{1}) + A r e a (A_{2}) - A r e a (A_{1} \cap A_{2})$
But, $A r e a (A_{1}) = A r e a (A_{2}) = 1.$ (as given in question)
Also, $A r e a (A_{1} \cap A_{2}) < \frac{1}{13}$ (as per our assumption)
Hence, $A r e a (A_{1} \cup A_{2}) > 1 + 1 - \frac{1}{13}$
$A r e a (A_{1} \cup A_{2}) > 2 - \frac{1}{13}$
Similarly,
$A r e a (A_{1} \cup A_{2} \cup A_{3}) = A r e a (A_{1}) + A r e a (A_{2}) + A r e a (A_{3})$
$- A r e a (A_{1} \cap A_{2}) - A r e a (A_{1} \cap A_{3}) - A r e a (A_{2} \cap A_{3})$
$+ A r e a (A_{1} \cap A_{2} \cap A_{3})$

But $A r e a (A_{1} \cap A_{2} \cap A_{3}) = 0$ (As per the condition in Question)
Thus, $A r e a (A_{1} \cup A_{2} \cup A_{3}) > 1 + 1 + 1 - \frac{1}{13} - \frac{1}{13} - \frac{1}{13} + 0$ .
i.e. $A r e a (A_{1} \cup A_{2} \cup A_{3}) > 3 - \frac{3}{13}$ .
Similarly,
$A r e a (A_{1} \cup A_{2} \cup A_{3} \cup A_{4}) = A r e a (A_{1}) + A r e a (A_{2}) + A r e a (A_{3}) + A r e a (A_{4})$
$- A r e a (A_{1} \cap A_{2}) - A r e a (A_{1} \cap A_{3}) - A r e a (A_{1} \cap A_{4})$
$- A r e a (A_{2} \cap A_{3}) - A r e a (A_{2} \cap A_{4}) - A r e a (A_{3} \cap A_{4})$
$+ A r e a (A_{1} \cap A_{2} \cap A_{3}) + A r e a (A_{1} \cap A_{2} \cap A_{4})$
$+ A r e a (A_{2} \cap A_{3} \cap A_{4}) - A r e a (A_{1} \cap A_{2} \cap A_{3} \cap A_{4})$

But $A r e a (A_{1} \cap A_{2} \cap A_{3} \cap A_{4}) = 0$
$A r e a (A_{1} \cap A_{2} \cap A_{3}) = 0$ , etc. (as per the condition in Question)
So, $A r e a (A_{1} \cup A_{2} \cup A_{3} \cup A_{4}) > 1 + 1 + 1 + 1 - 6 (\frac{1}{13}) + 3 (0) - 0$ .
i.e. $A r e a (A_{1} \cup A_{2} \cup A_{3} \cup A_{4}) > 4 - \frac{6}{13}$
In general,
$A r e a (A_{1} \cup A_{2} \cup A_{3} \cup . . . \cup A_{n}) > n - \frac{^{n} C_{2}}{13}$
Put $n = 13$ ;
$A r e a (A_{1} \cup A_{2} \cup A_{3} \cup . . . . \cup A_{13}) > 13 - \frac{^{13} C_{2}}{13}$
That is $A r e a (A_{1} \cup A_{2} \cup A_{3} \cup . . . . \cup A_{13}) > 7$
Hence, our assumption was wrong.
So, Area of overlap of any $2$ surfaces is not less than $\frac{1}{13}$ .

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13 Planar Surfaces, each of area 1 m2 are such that their total area is 7 m2. 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.

$13$ 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.