Chapter 14.4, Problem 37WE

a.

To determine

To find: What rotation maps $B$ to $A$ & $E$ to $D$ .

Answer to Problem 37WE

It is rotation $R_{C,90}$ for ${90}^{°}$ counter-clockwise about point $C$ .

Explanation of Solution

Given:

  $\Delta ABC$ and $\Delta DEC$ are isosceles right triangles.

Concept Used:

A rotation is a transformation suggested by rotating a paddle wheel. When wheel stops, new position of a paddle referred as the image of initial position.

Calculation:

As per the given problem

  

  $\Delta ABC$ and $\Delta DEC$ are isosceles right triangles.

It is rotation $R_{C,90}$ for ${90}^{°}$ counter-clockwise about point $C$ .

  

Hence,

It is rotation $R_{C,90}$ for ${90}^{°}$ counter-clockwise about point $C$ .

b.

To determine

To proof: $AD=BE$

Answer to Problem 37WE

  $AD=BE$

Explanation of Solution

Given:

  $\Delta ABC$ and $\Delta DEC$ are isosceles right triangles.

Concept Used:

A rotation is a transformation suggested by rotating a paddle wheel. When wheel stops, new position of a paddle referred as the image of initial position.

Calculation:

As per the given problem

  

  $\Delta ABC$ and $\Delta DEC$ are isosceles right triangles.

It is rotation $R_{C,90}$ for ${90}^{°}$ counter-clockwise about point $C$ , maps $B$ to $A$ & $E$ to $D$ .

  

Since rotation is isometry, we can say $AD=BE$

Hence,

  $AD=BE$

c.

To determine

To explain: $\overline{AD}\perp \overline{BE}$

Answer to Problem 37WE

  $\overline{AD}\perp \overline{BE}$

Explanation of Solution

Given:

  $\Delta ABC$ and $\Delta DEC$ are isosceles right triangles.

Concept Used:

A rotation is a transformation suggested by rotating a paddle wheel. When wheel stops, new position of a paddle referred as the image of initial position.

Calculation:

As per the given problem

  

  $\Delta ABC$ and $\Delta DEC$ are isosceles right triangles.

It is rotation $R_{C,90}$ for ${90}^{°}$ counter-clockwise about point $C$ , maps $B$ to $A$ & $E$ to $D$ .

  

Since $BE$ is rotated ${90}^{°}$ counter-clockwise about point $C$ , angle between $AD$ & $BE$ is ${90}^{°}$ .

Hence,

  $\overline{AD}\perp \overline{BE}$