Question

1. polygon ( abcde ) is congruent to polygon ( fghjk ). complete the sentences describing a sequence of rigid motions that takes polygon ( abcde ) to polygon ( fghjk ).

• translate polygon ( abcde ) so that ( b ) goes to point
• next, rotate the image (direction) around point ( g ) so that line ( bc ) coincides with line
• finally, reflect the image over line

1. triangle ( abc ) is congruent to triangle ( abc ). describe a sequence of rigid motions that takes ( a ) to ( a ), ( b ) to ( b ), and ( c ) to ( c ).
2. select all true statements about this figure...

a) ( y + z = z + x )
b) ( w + x = 180 )
c) reflect across line ( ae ). then angle ( adb ) is the image of angle ( cde ).
d) rotate ( 180^{circ} ) using the center ( d ). then the angle ( cda ) is the image of angle ( bde ).
e) rotate counterclockwise by angle ( edc ) using center ( d ). then angle ( adc ) is the image of ( bde ).
f) reflect across the perpendicular bisector of line ( ae ). then angle ( edc ) is the image of angle ( adb ).

Explanation:

1)

Step1: Translation

Since the polygons are congruent, we look at the corresponding vertices. For the translation, we want to map \(B\) to its corresponding vertex in the other polygon. By comparing the two polygons \(ABCDE\) and \(FGHJK\), the corresponding vertex of \(B\) is \(G\).

Step2: Rotation

After translating \(B\) to \(G\), we need to rotate. Let's assume the orientation. If we consider the sides, a clock - wise rotation (by observing the relative positions of the sides). After translation, to make \(BC\) coincide with its corresponding side. The corresponding side of \(BC\) is \(GH\).

Step3: Reflection

After translation and rotation, to make the two polygons coincide, we reflect over the line \(GH\) (or \(BC\)'s corresponding line after rotation).

2)

Step1: Translation

Translate triangle \(ABC\) so that \(A\) maps to \(A'\). Let the translation vector be \(\overrightarrow{AA'}\). So, \(T(x,y)=(x + a,y + b)\) where \(a=x_{A'}-x_A\) and \(b = y_{A'}-y_A\)

Step2: Rotation (if needed)

If after translation \(B\) is not at \(B'\) and \(C\) is not at \(C'\), we rotate the translated triangle around \(A'\). Let \(\theta\) be the angle between \(\overrightarrow{A'B}\) (after translation) and \(\overrightarrow{A'B'}\). We rotate the translated triangle by \(\theta\) around \(A'\)

Step3: Reflection (if needed)

If after translation and rotation the triangle is not congruent (in terms of orientation), we reflect the triangle over the line containing \(A'B'\)

3)

a)

Since \(y\) and \(x\) are not equal (in general, vertical - angle and adjacent - angle relationships). \(y+z\) and \(z + x\): \(y
eq x\) (unless \(x = y\) which is not given by the general intersection of lines). \(y+z
eq z + x\)

b)

\(w\) and \(x\) are adjacent angles forming a linear pair. By the linear - pair postulate, \(w + x=180^{\circ}\)

c)

When we reflect across line \(AE\), \(\angle ADB\) and \(\angle CDE\) are vertical angles. A reflection across \(AE\) (which is a line of symmetry in terms of the angle - pair relationship) will map \(\angle ADB\) to \(\angle CDE\)

d)

A \(180^{\circ}\) rotation about \(D\) is a central symmetry. \(\angle CDA\) and \(\angle BDE\) are vertical angles. A \(180^{\circ}\) rotation about \(D\) will map \(\angle CDA\) to \(\angle BDE\)

e)

If we rotate counter - clockwise by \(\angle EDC\) around \(D\), \(\angle ADC\) will not be the image of \(\angle BDE\). The rotation angle and the relationship between the angles do not match for this mapping.

f)

The perpendicular bisector of \(AE\) is a line of symmetry for the pair of angles \(\angle EDC\) and \(\angle ADB\). Reflecting across the perpendicular bisector of \(AE\) will map \(\angle EDC\) to \(\angle ADB\)

Answer:

1)

• Translate Polygon \(ABCDE\) so that \(B\) goes to point \(G\)
• Next, rotate the image (direction) clock - wise around point \(G\) so that line \(BC\) coincides with line \(GH\)
• Finally, reflect the image over line \(GH\)

1. Translate \(\triangle ABC\) by the vector \(\overrightarrow{AA'}\), then rotate (if necessary) around \(A'\) by the angle between \(\overrightarrow{A'B}\) (after translation) and \(\overrightarrow{A'B'}\), and reflect (if necessary) over the line containing \(A'B'\)

3)

• b) \(w + x = 180\)
• c) Reflect across line \(AE\). Then angle \(ADB\) is the image of angle \(CDE\)
• d) Rotate \(180^{\circ}\) using the center \(D\). Then the angle \(CDA\) is the image of angle \(BDE\)
• f) Reflect across the perpendicular bisector of line \(AE\). Then angle \(EDC\) is the image of angle \(ADB\)