Chapter 8.4, Problem 27WE

To determine

To find: $OB$ , $OC$ , $OD$ and $OE$ .

Answer to Problem 27WE

  $OB=\sqrt{2}$

  $OC=2$

  $OD=2\sqrt{2}$

  $OE=4$

Explanation of Solution

Given:

The figure:

All these four triangles are $45°-45°-90°$ .

Concept Used:

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

Consider the given figure:

It is given that the triangle $△OAB$ is a $45°-45°-90°$ .

Her $\overline{OB}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{OA}$ and $\overline{AB}$ are the legs.

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

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

  $\begin{array}{c}OB=\sqrt{2}\left(OA\right)\\ =\sqrt{2}\left(1\right)\\ =\sqrt{2}\end{array}$

Similarly, it is given that the triangle $△OBC$ is a $45°-45°-90°$ .

Her $\overline{OC}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{OB}$ and $\overline{BC}$ are the legs.

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

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

  $\begin{array}{c}OC=\sqrt{2}\left(OB\right)\\ =\sqrt{2}\left(\sqrt{2}\right)\\ =2\end{array}$

Similarly, it is given that the triangle $△OCD$ is a $45°-45°-90°$ .

Her $\overline{OD}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{OC}$ and $\overline{CD}$ are the legs.

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

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

  $\begin{array}{c}OD=\sqrt{2}\left(OC\right)\\ =\sqrt{2}\left(2\right)\\ =2\sqrt{2}\end{array}$

Similarly, it is given that the triangle $△ODE$ is a $45°-45°-90°$ .

Her $\overline{OE}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{OD}$ and $\overline{DE}$ are the legs.

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

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

  $\begin{array}{c}OE=\sqrt{2}\left(OD\right)\\ =\sqrt{2}\left(2\sqrt{2}\right)\\ =4\end{array}$