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How does a dilation change a figure, and how do you prove two figures similar?

Perform dilations on the coordinate plane and describe their effect on lengths and angles; define similarity through a sequence of rigid motions and a dilation; and prove triangles similar using AA (and SAS, SSS similarity), then use proportions to find missing lengths.

A NY Regents Geometry answer on dilations and similarity: performing a dilation about a center, why angles are preserved while lengths scale, the AA similarity criterion, and using proportions to find missing sides.

Generated by Claude Opus 4.89 min answer

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  1. What this topic is asking
  2. Dilations
  3. Similarity through transformations
  4. Proportions in similar figures
  5. Try this

What this topic is asking

The Regents Geometry exam (the Similarity, Right Triangles, and Trigonometry, G-SRT, cluster) wants you to perform a dilation, know that it scales lengths by the scale factor while keeping angles unchanged, define similarity as a rigid motion followed by a dilation, and prove triangles similar (mostly by AA) then solve proportions for missing lengths. Similarity is the bridge from congruence to trigonometry.

Dilations

A dilation resizes a figure from a fixed center by a scale factor kk.

If k>1k > 1 the figure enlarges; if 0<k<10 < k < 1 it shrinks. Crucially, a dilation multiplies lengths by kk but preserves angle measures, so the image has the same shape. A line through the center maps to itself; a line not through the center maps to a parallel line.

Similarity through transformations

The Regents defines similarity transformationally: two figures are similar if a sequence of rigid motions and a dilation maps one onto the other. This is why similar figures have equal corresponding angles (rigid motions and dilations preserve angles) and proportional corresponding sides (the dilation scales them by kk).

For triangles, the standard proof is AA: if two angles of one triangle equal two angles of another, the triangles are similar (the third angles must match because angles sum to 180 degrees). SAS similarity (two sides in proportion with the included angle equal) and SSS similarity (all three sides in proportion) are the other criteria.

Proportions in similar figures

Once similarity is established, corresponding sides are proportional, and you solve for a missing length by cross-multiplying. A clarifying point that prevents most errors is to match corresponding parts carefully: write the ratio so that the same triangle's sides are on top and the other triangle's sides are on the bottom (or numerator/denominator consistently), and pair sides that lie opposite equal angles. A second idea worth noting is that lengths scale by kk, but areas scale by k2k^2: doubling the sides of a figure multiplies its area by four, a relationship the Regents tests when comparing similar figures' areas.

Try this

Q1. A dilation with scale factor 12\frac{1}{2} maps a side of length 12 to what length? [1 credit]

  • Cue. Multiply by 12\frac{1}{2}: length 6.

Q2. Two triangles have angle pairs 40,7040, 70 and 40,7040, 70 degrees. Are they similar? [1 credit]

  • Cue. Two pairs of equal angles means AA similarity: yes.

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart I (multiple choice). Triangle ABCABC is dilated by a scale factor of 3 centered at the origin. If side AB=5AB = 5, what is the length of the corresponding side ABA'B'? (1) 55 (2) 88 (3) 1515 (4) 53\frac{5}{3}
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The correct answer is (3).

A dilation multiplies every length by the scale factor. With scale factor 3, the image side is 3×5=153 \times 5 = 15. Choice (2) wrongly adds 3 instead of multiplying; choice (4) divides, which would be a scale factor of 13\frac{1}{3}. Angles are unchanged by a dilation, but lengths scale by the factor.

Regents (style)4 marksPart III (constructed response). In triangle ABCABC, point DD is on AB\overline{AB} and point EE is on AC\overline{AC} with DEBC\overline{DE} \parallel \overline{BC}. Given AD=4AD = 4, DB=6DB = 6, and AE=5AE = 5, find ECEC and explain the similarity used.
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A 4-credit question: credit for the similarity statement, the proportion, and the answer.

Because DEBC\overline{DE} \parallel \overline{BC}, angle ADEADE \cong angle ABCABC and angle AEDAED \cong angle ACBACB (corresponding angles), so triangle ADEADE \sim triangle ABCABC by AA. Corresponding sides are proportional: ADAB=AEAC\frac{AD}{AB} = \frac{AE}{AC}. Here AB=AD+DB=10AB = AD + DB = 10 and AC=AE+EC=5+ECAC = AE + EC = 5 + EC. So 410=55+EC\frac{4}{10} = \frac{5}{5 + EC}, giving 4(5+EC)=504(5 + EC) = 50, so 20+4EC=5020 + 4EC = 50, 4EC=304EC = 30, EC=7.5EC = 7.5. A correct proportion with no AA justification, or setting up the ratio with the wrong corresponding parts, loses credit.

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