Chapter 4.4, Problem 29HP

(a)

To determine

To explain:

Two methods to use to prove that $\Delta WYZ$ is congruent to $\Delta WYX$ and to state the efficient method of the two.

Answer to Problem 29HP

Method 1: Using the Distance Formula to find the length of each side, and then using the Postulate of Side-Side-Side Congruence to show the congruent triangles.

Method 2: Find the slopes of $\overline{ZX}$ and $\overline{WY}$ to prove that they are perpendicular and that $\angle WYZ$ and $\angle WYX$ are both right angles. Thus, can use the Distance Formula to prove that $\overline{XY}$ is congruent to $\overline{ZY}$ .

The method 2 is more efficient, because only need two steps to solve instead of three.

Explanation of Solution

Given Information:

  

Method 1: Using the Distance Formula to find the length of each side, and then using the Postulate of Side-Side-Side Congruence to show the congruent triangles.

Method2: Find the slopes of $\overline{ZX}$ and $\overline{WY}$ toprove that they are perpendicular and that $\angle WYZ$ and $\angle WYX$ are both right angles. Thus, can use theDistance Formula to prove that $\overline{XY}$ is congruent to $\overline{ZY}$ .

Since the triangles share the leg $\overline{WY}$ , thus, can use the Side-Angle-Side Congruence Postulate.

Thus, the method 2 is moreefficient, because only need two steps to solve instead ofthree.

(b)

To determine

To explain:

If the $\Delta WYZ$ and $\Delta WYX$ congruent.

Answer to Problem 29HP

Yes, $\Delta WYZ$ is congruent to $\Delta WYX$ by the Side-Angle-Side Congruent Postulate, since $\overline{WY}$ is congruent to $\overline{WY}$ .

Explanation of Solution

Given Information:

  

The slope of $\overline{WY}$ is $-1$ andthe slope of $\overline{ZX}$ is $1$ , and $-1$ and $1$ are oppositereciprocals, so $\overline{WY}$ is perpendicular to $\overline{ZX}$ .

Sincethey are perpendicular, $\angle WYZ$ and $\angle WYX$ are both $90°$ . Using the Distance Formula, the length of $\overline{ZY}$ is $\sqrt{{\left(4-1\right)}^2+{\left(5-2\right)}^2}=3\sqrt{2}$ and the length of $\overline{XY}$ is $\sqrt{{\left(7-4\right)}^2+{\left(8-5\right)}^2}=3\sqrt{2}$ .

Since $\overline{WY}$ is congruent to $\overline{WY}$ , $\Delta WYZ$ is congruent to $\Delta WYX$ by the Side-Angle-Side Congruent Postulate.