Question

rotating $\triangle jkl$ by $180^\circ$ counterclockwise about the origin, we get its image $\triangle jkl$ as shown below.
note that $\triangle jkl$ has vertices $j(-3, 7)$, $k(-5, 3)$, and $l(-8, 2)$.
also, note that $\triangle jkl$ has vertices $j(3, -7)$, $k(5, -3)$, and $l(8, -2)$.
complete the following.
scratch area (not part of answer)
(a) suppose the angles below have the same measure. choose the correct angle measure. use the protractor provided, as necessary.
$m\angle jcj = m\angle kck = m\angle lcl =$ select
(b) suppose each pair of segments below have the same length.
find each length. give exact answers (not decimal approximations).
$cj = cj = \square$ units
$ck = ck = \square$ units
$cl = cl = \square$ units
(c) choose the correct pair of statements about the rotation.
not all angles formed by a point and its image, with the vertex at the center of rotation, are congruent.
each point on the original figure is the same distance from the center of rotation as its image.
not all angles formed by a point and its image, with the vertex at the center of rotation, are congruent.
not every point on the original figure is the same distance from the center of rotation as its image.
all angles formed by a point and its image, with the vertex at the center of rotation, are congruent.
not every point on the original figure is the same distance from the center of rotation as its image.
all angles formed by a point and its image, with the vertex at the center of rotation, are congruent.
each point on the original figure is the same distance from the center of rotation as its image.

Explanation:

Part (a)

Step1: Recall 180° rotation property

A 180° counterclockwise (or clockwise) rotation about the origin means the angle between a point, the origin, and its image is 180°. So \( m\angle JCJ' = m\angle KCK' = m\angle LCL' = 180^\circ \).

Step1: Use distance formula for \( CJ \)

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \( d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2} \). For \( C(0,0) \) and \( J(-3,7) \), \( CJ=\sqrt{(-3 - 0)^2+(7 - 0)^2}=\sqrt{9 + 49}=\sqrt{58} \). Since rotation preserves distance, \( CJ' = CJ=\sqrt{58} \).

Step2: Use distance formula for \( CK \)

For \( C(0,0) \) and \( K(-5,3) \), \( CK=\sqrt{(-5 - 0)^2+(3 - 0)^2}=\sqrt{25 + 9}=\sqrt{34} \). So \( CK' = CK=\sqrt{34} \).

Step3: Use distance formula for \( CL \)

For \( C(0,0) \) and \( L(-8,2) \), \( CL=\sqrt{(-8 - 0)^2+(2 - 0)^2}=\sqrt{64 + 4}=\sqrt{68} = 2\sqrt{17} \). So \( CL' = CL=\sqrt{68} \) (or \( 2\sqrt{17} \)).

• A 180° rotation is a rigid transformation. Rigid transformations preserve distances (so each point and its image are equidistant from the center) and preserve angle measures (so all angles formed by a point, center, and its image are congruent, each \( 180^\circ \)). The last option states both these properties correctly.

Answer:

\( 180^\circ \)

Part (b)