Chapter 14.6, Problem 28WE

(a)

To determine

To verify: The result $P{P}^{″}=2\cdot AB$ , and write the theorem of midpoints that suggest to prove it.

Explanation of Solution

Given information: The given mapping is $H_B\circ H_A:P\to P^{″}$ and the figure is shown below:

  

Calculation:

A segment joining the midpoints of two sides of a triangle called as the mid-segment of a triangle. According to this theorem, the mid-segment of a triangle joins the midpoints of two sides of a triangle in such a manner that its length is half the length of the third side of the triangle.

The mid-segment of line $P{P}^{\prime }$ is $A$ and the mid-segment of line $P^{\prime }{P}^{″}$ is $B$ . So, the mid-segment of the triangle is $AB$ and third side of the triangle is $P{P}^{″}$ . By the theorem,

  $P{P}^{″}=2\cdot AB$

Therefore, it is verified that $P{P}^{″}=2\cdot AB$ .

(b)

To determine

To find: Whether the result $Q{Q}^{″}=2\cdot AB$ is correct or not.

Answer to Problem 28WE

The result $Q{Q}^{″}=2\cdot AB$ is correct.

Explanation of Solution

Given information: Consider another point $Q$ and the image of this point under $H_B\circ H_A$ is $Q^{″}$ . The figure is shown below:

  

Calculation:

A segment joining the midpoints of two sides of a triangle called as the mid-segment of a triangle. According to this theorem, the mid-segment of a triangle joins the midpoints of two sides of a triangle are such a manner that its length is half the length of the third sides of the triangle.

Consider another point $Q$ and its image $Q^{″}$ under the condition $H_B\circ H_A$ to form a triangle. Similarly the mid-segment of line $Q{Q}^{\prime }$ be $A$ and the mid-segment of line $Q^{\prime }{Q}^{″}$ be $B$ . So, the mid-segment of the triangle is $AB$ and third side of the triangle is $Q{Q}^{″}$ . By the theorem,

  $Q{Q}^{″}=2\cdot AB$

Therefore, it is verified that $Q{Q}^{″}=2\cdot AB$ .

(c)

To determine

To find: Whether $PQ$ and $P^{″}{Q}^{″}$ are equal, and the type of transformation $H_B\circ H_A$ .

Answer to Problem 28WE

The measure of $PQ$ and $P^{″}{Q}^{″}$ are equal. The transformation $H_B\circ H_A$ is an isometry.

Explanation of Solution

Consider the following figure.

  

As the figure show that the composite of two given half-turns is a translation and translation is an isometry.

Therefore, the measure of $PQ$ and $P^{″}{Q}^{″}$ are equal. The transformation $H_B\circ H_A$ is an isometry.