Question

is pqr similar to vwx? justify your answer.
yes, because a rotation 90° clockwise around the origin followed by a reflection over the x - axis maps pqr onto vwx.
yes, because a rotation 90° clockwise around the origin followed by a reflection over the y - axis maps pqr onto vwx.
no, because ∠r and ∠x do not have the same measure.

Explanation:

Step1: Understand similarity and transformations

Similarity can be established if one figure can be mapped to another via rigid - motions (rotations, reflections, translations) or a combination of them. Rigid - motions preserve angles and side - length ratios.

Step2: Analyze the rotation and reflection

A rotation of \(90^{\circ}\) clockwise around the origin and then a reflection over the \(x\) - axis.
Let's assume a general point \((x,y)\) in \(\triangle PQR\). A \(90^{\circ}\) clockwise rotation around the origin transforms the point \((x,y)\) to \((y, - x)\) (using the rotation formula \(

$$\begin{pmatrix}x\\y\end{pmatrix}$$

\to

$$\begin{pmatrix}\cos(- 90^{\circ})&-\sin(-90^{\circ})\\\sin(-90^{\circ})&\cos(-90^{\circ})\end{pmatrix}$$

$$\begin{pmatrix}x\\y\end{pmatrix}$$

=

$$\begin{pmatrix}0&1\\ - 1&0\end{pmatrix}$$

$$\begin{pmatrix}x\\y\end{pmatrix}$$

=

$$\begin{pmatrix}y\\-x\end{pmatrix}$$

\)). Then a reflection over the \(x\) - axis (transformation \((a,b)\to(a, - b)\)) would transform \((y,-x)\) to \((y,x)\).
If we consider the vertices of \(\triangle PQR\) and \(\triangle VWX\), we can check the side - length ratios and angle measures. Since rigid - motions (rotation and reflection) preserve the shape (angle measures) and the ratio of side lengths (for similar triangles, if it's a congruence transformation (a special case of similarity with ratio \(1\))).

Step3: Analyze the angle - based option

If we assume the side - lengths and angles of the triangles. For two triangles to be similar, their corresponding angles must be equal. If we consider the claim “No, because \(\angle R\) and \(\angle X\) do not have the same measure” without proper side - length or other angle comparisons (since in a triangle, if two angles are equal, the third is also equal by the angle - sum property of triangles (\(A + B + C=180^{\circ}\))), a single - angle non - equality claim is not sufficient. But since we can map \(\triangle PQR\) to \(\triangle VWX\) via a \(90^{\circ}\) clockwise rotation and then \(x\) - axis reflection (rigid - motions), the triangles are congruent (a special case of similarity with a scale factor of \(1\)).

Answer:

Yes, because a rotation \(90^{\circ}\) clockwise around the origin followed by a reflection over the \(x\) - axis maps \(PQR\) onto \(VWX\).