Question

question 7 (multiple choice worth 1 points) (03.09 mc) jamal drew δlmn, in which m∠lmn = 90°. he then drew δpqr, which was a dilation of δlmn by a scale factor of 2 from the center of dilation at point m. which of these can be used to prove δlmn ~ δpqr by the aa similarity postulate?
m∠p ≅ m∠n; this can be confirmed by translating point p to point n
lm = 2pq; this can be confirmed translating point p to point l
m∠r ≅ m∠n; this can be confirmed by translating point r to point n
mn = 2qr; this can be confirmed translating point r to point n

Explanation:

To solve this, we analyze each option based on dilation properties and AA similarity (two angles congruent):

Step 1: Recall Dilation and AA Similarity

Dilation preserves angle measures (angles are congruent) and scales side lengths by the scale factor. AA similarity requires two pairs of congruent angles. $\triangle LMN$ has $\angle LMN = 90^\circ$, so $\triangle PQR$ (dilation of $\triangle LMN$) will also have a right angle, and we need one more pair of congruent angles.

Analyze Each Option:

• Option 1: $m\angle P \cong m\angle N$ (translate $P$ to $M$)

Dilation from $M$: $P$ corresponds to $L$ (not $M$ or $N$), so this is incorrect.

• Option 2: $\overline{LM} = 2\overline{PQ}$ (translate $P$ to $L$)

This is about side lengths, not angles. AA similarity needs angle congruence, so this is irrelevant.

• Option 3: $m\angle R \cong m\angle N$ (translate $R$ to $M$)

Dilation from $M$: $R$ corresponds to $N$ (since dilation center is $M$, $MN$ scales to $MR$? Wait, no—dilation center $M$: $M$ maps to $M$, $L$ maps to $P$, $N$ maps to $R$. So $\angle N$ (in $\triangle LMN$) corresponds to $\angle R$ (in $\triangle PQR$). Translating $R$ to $N$ confirms $\angle R \cong \angle N$. Also, $\angle LMN = \angle PQR = 90^\circ$ (dilation preserves right angle). Thus, two angles ($\angle LMN \cong \angle PQR$, $\angle N \cong \angle R$) are congruent, satisfying AA similarity.

• Option 4: $\overline{MN} = 2\overline{QR}$ (translate $R$ to $N$)

This is about side lengths, not angles. AA similarity needs angles, so this is irrelevant.

Answer:

The correct option is: $m\angle R \cong m\angle N$; this can be confirmed by translating point $R$ to point $N$ (the third option, depending on labeling—assuming the option with $m\angle R \cong m\angle N$ and translating $R$ to $N$).