Question

geometry mock ia exam directions: show all of your work for each question, making sure to draw and annotate all diagrams, and select the best matching answer. grade your work against the answer key to see how you did. once you see your score and the correct answers, begin completing corrections for the ones you got incorrect and redo them until they’re correct. this will be counted as a quiz grade. good luck! *#1.) given: \\(\overline{ae}\\) bisects \\(\overline{bd}\\) at \\(c\\) and \\(\angle abc \cong \angle edc\\) which statement is needed to prove \\(\triangle abc \cong \triangle edc\\) using asa? a. \\(\angle bca \cong \angle dce\\) b. \\(\angle dec \cong \angle bac\\) c. \\(\angle abc\\) and \\(\angle edc\\) are right angles d. \\(\overline{bd}\\) bisects \\(\overline{ae}\\) at \\(c\\) #2.) a regular polygon is rotated \\(216^\circ\\) around its center and is mapped onto itself. the figure could have how many sides? a. 3 b. 5 c. 11 d. 12 #3.) which of the following lines of reflections does not map square \\(abcd\\) above onto itself? a. \\(x = -1.5\\) b. \\(y = 0\\) c. \\(y = 1.5\\) d. \\(y = -x\\)

Explanation:

Question #1

To prove \(\triangle ABC \cong \triangle EDC\) using ASA (Angle - Side - Angle) congruence criterion, we need two angles and the included side to be congruent. We know that \(\overline{AE}\) bisects \(\overline{BD}\) at \(C\), so \(BC = DC\) (by the definition of a bisector). We are also given that \(\angle ABC\cong\angle EDC\). For ASA, we need the included angle between the side \(BC = DC\) and the given angle. \(\angle BCA\) and \(\angle DCE\) are vertical angles, and vertical angles are congruent. If we have \(\angle ABC\cong\angle EDC\), \(BC = DC\), and \(\angle BCA\cong\angle DCE\), then we can apply ASA. Option b is for AAS (Angle - Angle - Side) or would be used for a different congruence approach. Option c is not necessary as we don't need the angles to be right angles. Option d is about \(\overline{BD}\) bisecting \(\overline{AE}\), which is not related to the ASA criterion for these triangles.

Step 1: Recall the formula for rotational symmetry of a regular polygon

For a regular polygon with \(n\) sides, the minimum angle of rotation that maps the polygon onto itself is given by \(\theta=\frac{360^{\circ}}{n}\). If a rotation of \(216^{\circ}\) maps the polygon onto itself, then \(216^{\circ}\) must be a multiple of \(\frac{360^{\circ}}{n}\), or equivalently, \(n\) must be a factor of \(\frac{360^{\circ}}{gcd(216^{\circ}, 360^{\circ})}\). First, find the greatest common divisor (gcd) of \(216\) and \(360\). Using the Euclidean algorithm: \(360=216\times1 + 144\), \(216 = 144\times1+72\), \(144=72\times2 + 0\). So \(gcd(216, 360)=72\). Then \(\frac{360}{72} = 5\). We can also check by the formula \(\theta=\frac{360}{n}\), and we know that \(216 = k\times\frac{360}{n}\) for some integer \(k\). Rearranging, \(n=\frac{360k}{216}=\frac{5k}{3}\). Since \(n\) must be an integer, \(k\) must be a multiple of \(3\). Let \(k = 3\), then \(n = 5\). Let's verify: for \(n = 5\), the minimum rotation angle is \(\frac{360}{5}=72^{\circ}\), and \(216\div72 = 3\), so a rotation of \(3\times72^{\circ}=216^{\circ}\) will map the pentagon onto itself. For \(n = 3\), the minimum rotation angle is \(120^{\circ}\), \(216\div120 = 1.8\) (not an integer). For \(n = 11\), the minimum rotation angle is \(\frac{360}{11}\approx32.73^{\circ}\), \(216\div\frac{360}{11}=216\times\frac{11}{360}=\frac{66}{10} = 6.6\) (not an integer). For \(n = 12\), the minimum rotation angle is \(30^{\circ}\), \(216\div30 = 7.2\) (not an integer).

Step 2: Conclusion

So the polygon could have \(5\) sides.

A square has reflection symmetry over its vertical mid - line, horizontal mid - line, and the lines \(y = x\) and \(y=-x\) (depending on the position of the square). Looking at the square \(ABCD\) (from the diagram, we can assume the square is centered or has a certain position). The line \(y = 0\) (the x - axis) does not map the square onto itself. The line \(x=- 1.5\) could be the vertical mid - line, \(y = 1.5\) could be the horizontal mid - line, and \(y=-x\) is a line of symmetry for a square (if the square is positioned appropriately). The line \(y = 0\) (x - axis) will not reflect the square onto itself as the square is above or below the x - axis (from the diagram's appearance with points \(B\), \(A\), \(C\), \(D\) which seem to be in a non - x - axis symmetric position for \(y = 0\)).

Answer:

a. \(\angle BCA \cong \angle DCE\)

Question #2