Chapter 8.4, Problem 7WE

To determine

To complete: the given table.

Answer to Problem 7WE

a	b	c
$4\sqrt{2}$	$4\sqrt{2}$	$8$

Explanation of Solution

Given:

The figure:

And the table:

a	b	c
?	$4\sqrt{2}$	?

Concept Used:

Theorem 3-11: The addition of measures of all angles of a triangle is 180.

Theorem 8-6: in a $45°-45°-90°$ triangle, the hypotenuse is $\sqrt{2}$ times as long as legs of triangle.

Label the given figure as shown below:

Apply theorem 3-11 on the triangle $△ABC$ ,

  $\begin{array}{c}m\angle A+m\angle B+m\angle C=180\\ m\angle A+90+45=180\\ m\angle A+135=180\end{array}$

  $\begin{array}{c}m\angle A=180-135\\ m\angle A=45\end{array}$

Thus, the given triangle is a $45°-45°-90°$ . Here c is the side opposite to the right angle, so it is the hypotenuse and a and b are the legs.

Now, a $45°-45°-90°$ is an isosceles triangle and thus both its legs are congruent.

Thus,

  $a=b=4\sqrt{2}$

Now, by theorem 8-6, the hypotenuse is $\sqrt{2}$ times as long as a leg, so

  $\begin{array}{c}c=\sqrt{2}a\\ =\sqrt{2}\left(4\sqrt{2}\right)\\ =4\times 2\\ =8\end{array}$

Thus, the complete table is:

a	b	c
$4\sqrt{2}$	$4\sqrt{2}$	$8$