Acomputer has been programmed to draw a rectangle with a perimeter of 20 cm. the program randomly chooses integer lengths. what is the expected area of the rectangle, to the nearest hundredth?

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:

Where is the length and is the width.

Therefore,

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:

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:

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.

side side side

side angle angle

angle side angle

side angle side

step-by-step explanation:

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)