If square pqrs is rotated 180° about the origin to create square p'q'r's', which set of sides would be parallel in the resulting image?

i. p'q' and q'r'
ii. q'r' and p's'
iii. p'q' and s'p'
iv. q'r' and r's'

a.
ii only
b.
iii only
c.
i and iv only
d.
i and ii only

A. II only is correct.

Step-by-step explanation:

We are given the square PQRS with co-ordinates P(2,-5), Q(4,-3), R(6,-5) and S(4,-7).

As, we know that,

Reflection about 180° changes the co-ordinates (x,y) to (-x,-y).

So, the co-ordinates of P'Q'R'S' are given by,

P'(-2,-5), Q'(-4,-3), R'(-6,-5) and S'(-4,-7).

So, from the figure below, we see that,

P'Q' is parallel to S'R'

Q'R' is parallel to P'S'.

So, according to the options, we have,

Option II i.e. Q'R' is parallel to P'S' is correct.

Thus, option A is correct.

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3