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Viefleur [7K]
4 years ago
6

Do you guys know what’s the answer

Mathematics
1 answer:
larisa [96]4 years ago
8 0
You can see reflections everywhere. In <em>mirror, glass, lake, river, beach, etc.</em> So if you take a <em>central line</em> every point is the same distance from the central line (also called <em>mirror line</em> or <em>axis of symmetry</em>). So let's analyze each item in this problem to find what is true.

So we have the triangle ΔABC and the line of reflection \overleftrightarrow{PT}, that is, the mirror line. We also know the reflected object that is ΔA'B'C', thus:

1. A'A=C'C

This is false. The distance from A to the mirror line is equal to the distance from this line to the point A'. On the other hands, the distance from C to the mirror line is equal the distance from this line to the point C' but it doesn't mean that necessarily A'A=C'C. You might have the point A far away from the line while the point C might be near to it, or vice versa, giving place to different distances.

2. C'Q=QC

This is true. <span>The main characteristic of reflecting shapes is that every point of it is <em>the same distance from the central line. </em>

3. </span>\overleftrightarrow{PT} \perp \overline{A'A}

This is true. When reflecting points, you <em>must</em> reflect them across the mirror line <em>always perpendicularly.</em> So the given segment is perpendicular to the line.

4. \overline{C'C}=\overline{B'B}

This is false. In fact, this two lines are parallel.

5. \overline{A'A} \parallel \overline{B'B}

This is true given that every reflecting point is reflected across the line <em>perpendicularly</em> to it.

6. m \angle TRB=90^{\circ}
<span>
This is true. Given that the segment \overline{B'B} is perpendicular to the mirror line, then this angle must be equal to 90^{\circ}
 </span>
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The correct question is
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