Chapter 9.1, Problem 7P

To determine

To prove: The sum of the lengths of the sides of a parallelogram is equal to the sum of the squares of the lengths of its diagonals.

Explanation of Solution

Given information: The points are $C\left(0,0\right),D\left(5,0\right),E\left(8,5\right)$ and $F\left(3,5\right)$ .

Formula used: $D=\sqrt{{\left(c-a\right)}^2+{\left(d-b\right)}^2}$

Proof: Apply the formula to sides lengths between the given points,

  $\begin{array}{l}\Rightarrow CD=\sqrt{{\left(5-0\right)}^2+{\left(0-0\right)}^2}\\ =5\\ \Rightarrow DE=\sqrt{{\left(8-5\right)}^2+{\left(5-0\right)}^2}\\ =\sqrt{9+25}\\ =\sqrt{34}\end{array}$

Similarly,

  $\begin{array}{l}\Rightarrow EF=\sqrt{{\left(3-8\right)}^2+{\left(5-5\right)}^2}\\ =5\\ \Rightarrow FC=\sqrt{{\left(3-0\right)}^2+{\left(5-0\right)}^2}\\ =\sqrt{9+25}\\ =\sqrt{34}\end{array}$

Hence, the sum of the square of the sides,

  $25+34+25+34=118$

Now, calculate the diagonals length,

  $\begin{array}{l}\Rightarrow CE=\sqrt{{\left(8-0\right)}^2+{\left(5-0\right)}^2}\\ =\sqrt{64+25}\\ =\sqrt{89}\\ \Rightarrow DF=\sqrt{{\left(5-0\right)}^2+{\left(3-5\right)}^2}\\ =\sqrt{25+4}\\ =\sqrt{29}\end{array}$

Hence, the sum of the squares of the diagonals,

  $89+29=118$

Hence, the sum of the lengths of the sides of a parallelogram is equal to the sum of the squares of the lengths of its diagonals.

Hence, proved.