Chapter 8.4, Problem 4ST1

To determine

To find: the length of the diagonal

Answer to Problem 4ST1

  $AC=4\sqrt{5}$

Explanation of Solution

Given:

The rectangle has length 8 and width 4.

Concept Used:

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.

Draw a rectangle of length 8 and width 4 and label it as $ABCD$ .

And let the length of the diagonal $\overline{AC}$ be x.

Consider the right triangle $△ADC$ .

Now, $\overline{AC}$ is the hypotenuse and has length x and the lengths of the legs are 8 and 4.

So, by theorem 8-2,

  $\begin{array}{c}A{C}^2=A{D}^2+D{C}^2\\ x^2=4^2+8^2\\ x^2=16+64\\ x^2=80\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{x^2}=\sqrt{80}\\ x=4\sqrt{5}\end{array}$

So, the length of the diagonal of the given triangle is $4\sqrt{5}$.