Chapter 8.4, Problem 33E

(a)

To determine

To find: The perimeter and the area of the square.

Answer to Problem 33E

The perimeter and the area of the square are $16\text{}\text{unit}$ and $16\text{sq}\text{.}\text{}\text{unit}$

Explanation of Solution

Given information: The diagram is given below.

  

Formula used: $P=4s$ and $A=s^2$

Calculation: From the diagram,

  $\begin{array}{l}LM=\left|2-\left(-2\right)\right|\\ =\left|2+2\right|\\ =\left|4\right|\\ \therefore LM=4\end{array}$

  $\begin{array}{l}MN=\left|-2-2\right|\\ =\left|-4\right|\\ \therefore MN=4\end{array}$

Therefore, the area will be

  $\begin{array}{l}A=s^2\\ A=4^2\\ A=16\text{sq}\text{.}\text{}\text{unit}\end{array}$

Similarly, the perimeter,

  $\begin{array}{l}P=4s\\ P=4\times 4\\ P=16\text{}\text{unit}\end{array}$

(b)

To determine

To Explain: The joining midpoints of the square make a quadrilateral.

Answer to Problem 33E

The formed quadrilateral is a square.

Explanation of Solution

Given information: The diagram is given below.

  

Formula used: $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Calculation: Get the mid-points and then connect them,

  $\begin{array}{l}\Rightarrow LM=\left(\frac{-2+2}{2},\frac{2+2}{2}\right)\\ LM=\left(0,2\right)\\ \Rightarrow MN=\left(\frac{2+2}{2},\frac{2-2}{2}\right)\\ MN=\left(2,0\right)\end{array}$

  $\begin{array}{l}\Rightarrow NP=\left(\frac{-2+2}{2},\frac{-2-2}{2}\right)\\ NP=\left(0,-2\right)\\ \Rightarrow PL=\left(\frac{-2-2}{2},\frac{-2+2}{2}\right)\\ PL=\left(-2,0\right)\end{array}$

Therefore, the distance of the midpoints,

  $\begin{array}{l}=\sqrt{{\left(-2-0\right)}^2+\left(0-2\right)}\\ =\sqrt{4+4}\\ =\sqrt{8}\\ =2\sqrt{2}\end{array}$

The formed quadrilateral is also a square because each sides of quadrilateral are equal.

  

(c)

To determine

To Explain: The area and the perimeter of the formed quadrilateral and compare with area of part (a).

Answer to Problem 33E

The area and the perimeter of the formed quadrilateral are $8\sqrt{2}\text{unit}$ and $8\text{sq}\text{.}\text{unit}$ .

Explanation of Solution

Given information: The diagram is given below.

  

Formula used: $P=4s$ and $A=s^2$

Calculation: The area of the quadrilateral will be,

  $\begin{array}{l}A=s^2\\ A={\left(2\sqrt{2}\right)}^2\\ A=8\text{sq}\text{.}\text{units}\end{array}$

The perimeter of the quadrilateral will be,

  $\begin{array}{l}P=4s\\ P=4\times 2\sqrt{2}\\ P=8\sqrt{2}\text{units}\end{array}$

The area and the perimeter of the formed quadrilateral are $8\sqrt{2}\text{unit}$ and $8\text{sq}\text{.}\text{unit}$ .

Therefore, both area and the perimeter of the formed inner square are approx half of the given square.