Chapter 15.3, Problem 4PSA

To determine

To Find: comparison between AB and XZ.

Answer to Problem 4PSA

  $AB<XZ$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Two triangle }\Delta \text{ABC and }\Delta \text{XYZ}\\ \to BC=7\\ \to AC=9\\ \to \angle C={75}^o\\ \to YZ=7\\ \to XY=9\\ \to \angle Y={80}^o\end{array}$

Concept:

The Hinge Theorem: It states that if two sides of two triangles are congruent and angle of one is greater than the angle of second then the third side of the triangle will be greater than the another.

Calculation:

  $\begin{array}{l}\to BC=7\\ \to AC=9\\ \to \angle C={75}^o\\ \to YZ=7\\ \to XY=9\\ \to \angle Y={80}^o\\ \text{⇒}BC\cong YZ\\ \text{⇒}AC\cong XY\\ \text{⇒}\angle Y>\angle C\\ \text{⇒}AB<XZ\left(\text{using SAS inequality or Hinge theorem}\right)\end{array}$

Conclusion:

Thus thecomparison shows that $AB<XZ$