Chelsea and Anika are trying to determine whether ΔABC and ΔPRQ can be proven congruent through rigid motions.

Chelsea says that ΔABC ≅ ΔPRQ because ΔABC can be rotated 270° clockwise to create ΔPRQ.

Anika says that ΔABC ≅ ΔPRQ because ΔABC can be reflected over the x-axis to create ΔPRQ.

a coordinate grid with triangles ABC and PQR at A negative 7 comma 2, B negative 3 comma 6, C negative 2 comma 1, P negative 7 comma negative 2, Q negative 2 comma negative 1, and R negative 3 comma negative 6

Who is correct?

Chelsea only
Anika only
Both Chelsea and Anika
Neither Chelsea nor Anika

Anika is correct

Step-by-step explanation:

* Lets revise the rules of reflection and rotation

- If point (x , y) reflected across the x-axis  then its image is (x , -y)

- If point (x , y) rotated about the origin by angle 270° clock wise  then

 its image is (-y , x)

* Lets check the vertices of the two triangles

- The vertices of Δ ABC are:

# A (-7 , 2)

# B (-3 , 6)

# C (-2 , 1)

- The vertices of Δ PRQ

# P (-7 , -2)

# R (-3 , -6)

# Q (-2 , -1)

* By comparing between the vertices of the two triangles

∵ Each y-coordinates of Δ PRQ has opposite sign of each

  y-coordinates of Δ ABC

∵ All x-coordinates of Δ PRQ are the same with x-coordinates of

 Δ ABC

- That means Δ PRQ is the image of Δ ABC after reflection across

 the x-axis  

∵ Reflection doesn't change the shape and the size of the figure

∴ Δ ABC and Δ PRQ have same size (equal sides and equal angles)

∴ Δ ABC is congruent to Δ PRQ

* Anika is correct

Anika

Step-by-step explanation:

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length