Please find the attached table to get a better understanding of the solution and reasoning provided here.
The perimeter of the randomly generated rectangle is 20 cm, thus:
[tex] 2(l+w)=20 [/tex]
Where [tex] l [/tex] is the length and [tex] w [/tex] is the width.
Therefore, [tex] l+w=10 [/tex]
As can be seen from the attached diagram, there are only four possibilities for the length and the breadth to have. The diagram also shows the area in each case.
Now, we will use the formula for the expected value of the area as:
[tex] EV=\sum A\times p [/tex]
Where EV is the Expected Value
A is the area
p is the probability of occurrence of the event where A will be the Area
Before we proceed, let us understand the table that has been attached.
Each column tells us the combination of length and breadth that we can have and also the corresponding area. For example the for a perimeter of 10, the integral value of length in the first column has been taken as 9, which obviously means that the breadth in that rectangle will be 1. Thus, making the area 9 (9 x 1).
Now, the probability of such an even happening is, obviously (as can be seen from the attached table), is 1/4. Thus, the expected value will be:
[tex] EV=\frac{1}{4}\times 9 + \frac{1}{4}\times 16+\frac{1}{4}\times 21+\frac{1}{4}\times 24=17.5 [/tex]
17.5 is the required answer.
Please note that the option for generation of a square is note taken into consideration because the question clearly mentions that the generated figure is a rectangle.
