Chapter 5.3, Problem 41PPS

To determine

To list: The sides of triangle from smallest to largest.

Answer to Problem 41PPS

The sides of triangle from smallest to largest is $AB<BC<AC<CD<BD$ .

Explanation of Solution

The measure of $\angle ACB$ is calculated as,

  $\begin{array}{c}m\angle ACB=180°-62°-71°\\ =47°\end{array}$

The measure of $\angle BCD$ is calculated as,

  $\begin{array}{c}m\angle BCD=180°-83°-42°\\ =55°\end{array}$

The measure of $\angle ACB$ is calculated as,

  $\begin{array}{c}m\angle ACB=180°-62°-71°\\ =47°\end{array}$

In $\Delta ABC$ .

  $m\angle ACB<m\angle A<m\angle ABC$

By Angle side relationship in triangles theorem.

The side from shortest to longest is $\overline{AB}<\overline{BC}<\overline{AC}$ .

In $\Delta BCD$ .

  $m\angle D<m\angle CBD<m\angle BCD$

By Angle side relationship in triangles theorem.

The side from shortest to longest is $\overline{BC}<\overline{CD}<\overline{BD}$ .

Also,

  $\begin{array}{c}AC<CD\\ 10<12\end{array}$

Thus,

  $AB<BC<AC<CD<BD$

Therefore, the sides of triangle from smallest to largest is $AB<BC<AC<CD<BD$ .