Chapter 8.4, Problem 33WE

To determine

To find: As many lengths as possible

Answer to Problem 33WE

  $\begin{array}{l}GJ=6\sqrt{3}\\ GI=6\\ GH=6\end{array}$

  $\begin{array}{l}HI=6\sqrt{2}\\ JH=12\end{array}$

Explanation of Solution

Given:

The figure:

Concept Used:

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

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

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

Theorem 8-7: In a $30°-60°-90°$ triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is $\sqrt{3}$ times as long as the shorter leg.

Consider the given figure:

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

  $\begin{array}{c}m\angle HIG+m\angle IGH+m\angle GHI=180\\ m\angle HIG+90+45=180\\ m\angle HIG+135=180\end{array}$

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

Similarly, Apply theorem 3-11 on the triangle $△JGI$ ,

  $\begin{array}{c}mJIG+m\angle IGJ+m\angle GJI=180\\ m\angle JIG+90+30=180\\ m\angle JIG+120=180\end{array}$

  $\begin{array}{c}m\angle JIG=180-120\\ m\angle JIG=60\end{array}$

Also the triangle $△JGI$ is a $30°-60°-90°$ . Here $\overline{JI}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{GI}$ is the shorter leg and $\overline{GJ}$ are the longer leg.

Now, by theorem 8-7, the hypotenuse is twice as long as the shorter leg, so

  $\begin{array}{c}JI=2GI\\ 12=2GI\\ \frac{12}{2}=GI\\ 6=GI\end{array}$

Also, by theorem 8-7, the longer leg is $\sqrt{3}$ times as long as the shorter leg.

Thus,

  $\begin{array}{c}GJ=\sqrt{3}GI\\ =\sqrt{3}\left(6\right)\\ =6\sqrt{3}\end{array}$

It can be observed that the triangle $△HGI$ is a $45°-45°-90°$ .

Her $\overline{HI}$ is the side opposite to the right angle, so it is the hypotenuse and $\overline{GH}$ and $\overline{GI}$ are the legs.

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

  $GH=GI=6$

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

  $\begin{array}{c}HI=\sqrt{2}\left(GI\right)\\ =\sqrt{2}\left(6\right)\\ =6\sqrt{2}\end{array}$

Consider the right triangle $△HGJ$ .

Now, $\overline{JH}$ is the hypotenuse and the lengths of the legs are $GH=6$ and $GJ=6\sqrt{3}$ .

So, by theorem 8-2,

  $\begin{array}{c}J{H}^2=G{H}^2+G{J}^2\\ =6^2+{\left(6\sqrt{3}\right)}^2\\ =36+108\\ =144\end{array}$

Now, take square root of both sides,

  $\begin{array}{c}\sqrt{J{H}^2}=\sqrt{144}\\ JH=12\end{array}$