Chapter 8.4, Problem 5ST1

To determine

To find: perimeter of square.

Answer to Problem 5ST1

  $20\sqrt{2}\text{ cm}$

Explanation of Solution

Given:

The diagonals of a square are of lengths 10 cm.

Concept Used:

Theorem 5-13: Diagonals of a square are perpendicular bisectors.

Theorem 8-2: the square of the hypotenuse of right triangle is equal to the sum of the squares of the side in a right triangle.

  $\text{Perimeter of square}=4\times \text{side}$

Draw a square with diagonals of lengths 10 cm.

Consider the right triangle $△ADC$ .

Now, $\overline{AC}$ is the hypotenuse and has length 10 and the lengths of the sides are x and x.

So, by theorem 8-2,

  $\begin{array}{c}A{C}^2=A{D}^2+D{C}^2\\ {10}^2=x^2+x^2\\ 100=2x^2\end{array}$

  $\begin{array}{c}{x}^2=\frac{100}{2}\\ x^2=50\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{50}\\ x=5\sqrt{2}\end{array}$

The length of side of square is $5\sqrt{2}$ .

Now, perimeter of square is,

  $\begin{array}{c}\text{Perimeter of square}=4\times \text{side}\\ =4\times 5\sqrt{2}\\ =20\sqrt{2}\end{array}$