Question

In△XYZ , XY¯¯¯¯¯¯=7in , YZ¯¯¯¯¯=5in , and XZ¯¯¯¯¯=4in . This triangle is reflected across the x-axis to result in △X'Y'Z' . Which is true? A X'Z'¯¯¯¯¯¯¯=2in B X'Y'¯¯¯¯¯¯¯=7in C Y'Z'¯¯¯¯¯¯¯=10in D Perimeter of △X'Y'Z'=32in

Answer 1

Answer:

B

Step-by-step explanation:

Reflection is a transformation that preserves lengths.

Triangle XYZ, with side lengths XY = 7 in, YZ = 5 in, XZ = 4 in, is reflected across the x-axis to result in △X'Y'Z'.

This means that corresponding sides have the same lengths:

• XY = X'Y' = 7 in;
• XZ = X'Z' = 4 in;
• YZ = Y'Z' = 5 in.

Find the perimeters of both triangles:

$P_{XYZ}=XY+XZ+YZ=7+4+5=16\ in\\ \\P_{X'Y'Z'}=X'Y'+X'Z'+Y'Z'=7+4+5=16\ in$

Hence, only option B is true.