Geometry Quiz

The image of ΔDEF is ΔD'E'F'. Under which transformation will the triangles not be congruent?

• a reflection through the origin

• a reflection over the line y = x

• a dilation with a scale factor of 1 centered at (2,3)

• dilation with a scale factor of 3/2 centered at the origin

In the diagram below, AD intersects BE at C, and AB || DE. If CD = 6.6 cm, DE = 3.4 cm, CE = 4.2 cm, and BC = 5.25 cm, what is the length of AC, to the nearest hundredth of a centimeter?

• 2.70

• 3.34

• 5.28

• 8.25

In the diagram below, triangle ACD has points B and E on sides AC and AD, respectively, such that BE || CD, AB = 1, BC = 3.5, and AD = 18. What is the length of AE, to the nearest tenth?

In the diagram below, ∠GRS ≅ ∠ART, GR = 36, SR = 45, AR = 15, and RT = 18. Which triangle similarity statement is correct?

• ΔGRS ~ ΔART by AA.

• ΔGRS ~ ΔART by SAS.

• ΔGRS ~ ΔART by SSS.

• ΔGRS is not similar to ΔART.

Use the diagram to answer the following question.

• A. 4.5

• B. 5

• C. 3

• D. 6

Triangle A'B'C' is the image of Triangle ABC after a dilation followed by a translation. Which statement(s) would always be true with respect to this sequence of transformations?

• A. II, only

• B. I and II

• C. II and III

• D. II, III, and IV

In the diagram below, AC = 7.2 and CE = 2.4. Which statement is not sufficient to prove \( \triangle ABC \cong \triangle EDC \)?

• A. \( \overline{AB} \) || \( \overline{ED} \)

• B. DE = 2.7 and AB = 8.1

• C. CD = 3.6 and BC = 10.8

• D. DE = 3.0, AB = 9.0, CD = 2.9, and BC = 8.7

In right triangle ABC shown below, altitude CD is drawn to hypotenuse AB. Explain why triangle ABC is similar to triangle ACD.

On the graph below, point A(3, 4) and line segment BC with coordinates B(4, 3) and C(2, 1) are graphed. What are the coordinates of B' and C' after line segment BC undergoes a dilation centered at point A with a scale factor of 2?

• B'(5, 2) and C'(1, –2)

• B'(6, 1) and C'(0, –1)

• B'(5, 0) and C'(1, –2)

• B'(5, 2) and C'(3, 0)

In the diagram below, line segment GI is parallel to line segment NT, and line segment IN intersects line segment GT at A. Prove: triangle GIA is similar to triangle TNA.