Geometry - 1st, 2nd & 3rd blocks

Your Geometry End of Course test is Monday, 12/3 and Tuesday, 12/4.  If you missed any of the answers to the EOC Review packet, the answers are below:

Unit 1:
1.    B      2.   D       3.  A       4.   C        5.   B         6.   C        7.   A        8.   D         9.   B   
Unit 2:
1.   D      2.  D        3.   A       4.  B         5.  C          6.   C        7.    C       8.   C         9.   D    
10.  Yes, triangle MPN is congruent to triangle ACB.  Segment MP is the same length as segment AC.   Segment NP is the same length as segment BC.  The included angles (angle P and angle C) are both right angles, so they are congruent.  So, the triangles are congruent by SAS.  You could also show that segments MN and AB are congruent (use the distance formula to find the lengths).  Then you could also say the two triangles are congruent by SSS.
Unit 3:
1.   A       2.  A        3.   C        4.   D        5.  B          6.   B        7.  B         8.  C            9.  A
10.  Ricardo should use tan to solve for the building height since he knows the length of the adjacent side and is trying to solve for the opposite side.  His equation would be tan 32 = (x / 75).  Using the chart, tan 32 = 0.6246, so substitute the decimal equivalent into the equation to get:   0.6246 = (x / 75).   To get x by itself, multiply both sides by 75.  x = (75)(0.6246).   The building is 46.8675 feet tall.
Unit 4:
1.    D        2.   B        3.  C         4.   D         5.  B          6.  D         7.   A          8.   B         9.    A   
10.  To find the area of one section, use the area of a sector formula.  You will need the radius and the angle measure.  The diameter is 12, so the radius is 6.  The circle is divided into 8 equal sections, so divide the entire circle (360) by the number of equal sections (8).   360 / 8 = 45 degrees.  To calculate the area of one section:  
((pi)(radius squared)(theta)) / 360
((3.14)(36)(45) / 360 ) = 14.13 square feet
Unit 5:
1.   C         2.   A        3.   B         4.   C           5.  C           6.  A          7.  A         8.   B          9.   D   
10.  You can approach this a few different ways.  The easiest is to first show that the figure has 4 right angles.   If the segments are perpendicular, then they meet at a 90 degree angle.  If the slopes of the segments are opposite reciprocals, then the segments are perpendicular (form a right angle).  The slopes of AB and DC are -3 and the slopes of CB and DA are 1/3.  Consecutive segments have slopes that are opposite reciprocals, so this figure has 4 right angles.  At this point you know it's either a square or a rectangle.  You need one additional piece of information to prove that it is a rectangle.  You can either prove that the diagonals are NOT perpendicular (the slopes of the diagonals are NOT opposite reciprocals) OR you can prove that the lengths of consecutive sides are NOT the same (use distance formula to find the lengths of the sides).
Unit 6:
1.   B          2.   A         3.   D         4.    D         5.   C          6.  C.         7.  C         8.    B          9.   A  
10.  When you roll two 6-sided number cubes to get the sum, there are 36 possibilities.  The given statement tells us that this is a conditional probability and that we are only looking at a subset of the population.  "Given that at least one of the rolls is a 3" limits our population to the 11 possible outcomes, so this is our denominator.  "The probability that the sum of the two rolls is a prime number" tell us the prime numbers from our subset is the numerator.  There are 4 prime numbers in our subset (5, 5, 7, 7).  So the probability is 4/11 or .36 or 36%.