*The following problems were taken from the*

**GEOMETRY (COMMON CORE)**Regents Exam given on Thursday, January 26, 2017.Previous problems can be found here.

### Part 1

**5. In the diagram below, if triangle ABE = triangle and AEFC is drawn, then
it could be proven that quadrilateral ABCD is a
**

(4) ** Parallelogram**.

First: common sense. If the figure could be proved to be a rectangle or a rhombus, it would have to be a parallelogram as well. You can't have multiple answers. If it were a square, then all four would be true.

Second: figures are not necessarily drawn to scale (even if they don't tell you), so you shouldn't assume that it isn't a square based

*solely*on the picture. Looking at an image isn't

*proof*.

If the triangles are congruent, then by **CPCTC** *(Corresponding parts of congruent triangles are congruent)* we know that AB = CD. We also know, for the same reason, that angle BAE = DCF. Those two angles are *alternate interior angles* along the transversal. That makes AB || CD.

If two sides of a quadrilateral are both parallel and congruent, the shape is a parallelogram.

We do not have any additional information to prove (nor *assume!*) that the shape is a rhombus.

**6. Under which transformation would triangle A'B'C', the image of triangle ABC, not be congruent to ABC?
**

(4) *dilation with a scale factor of 2 centered at the origin.*

A dilation increased the size, so it will no longer be congruent.

Continue to the next problems.

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