Since the measure of angle BAC is 42, the other two equal angles of triangle ABC must be (180 â 42)/2 =
138/2 = 69 degrees each. We will introduce the point F, which is the center of the circle, and draw segments from each vertex
to F. Since segment ED is tangent to the circle at point C, it must be perpendicular to radius CF. Angle BAC is bisected by
segment AF, so angle FAC is 21 degrees. Angle FCA is also 21 degrees since triangle AFC is isosceles. Thus, the measure of
angle ACD is 90 â 21 = 69 degrees, which is the same as the two base angles of triangle ABC.
Problem 6. To î¿nd the sum of the positive odds less than 100, we imagine adding 1 and 9, 3 and 97, 5 and 95, etc. In all, we can make 25 pairs that sum
to 100, which is 2500. The positive, two-digit multiples of 10 contribute another 450, and the sum of 4, 16 and 64 is 84. In all, we have 2500 +
450 + 84 = 3034. There are 50 + 9 + 3 = 62 numbers in our data set, so the mean (average) is 3034 Ã· 62 = 48.935 to the nearest thousandth. The median
of the odds less than 100 is 50. When the two-digit multiples of 10 are added, the median is still 50. However, when 4 and 16 are added on the lesser side
and 64 is added on the greater side, the median of the 62 numbers becomes the average of 50 and 49, which is 49.5. The positive difference between the
median and the mean of the set of numbers is thus 49.5 â 48.935 = 0.565.
Problem 7. Ryosuke traveled a distance of 74,592 â 74,568 = 24 miles between the time he picked up his friend and when he dropped him off. Since his
car gets 28 miles per gallon, he used 24/28 or 12/14 of a gallon. At $4.05 per gallon, the cost of the trip is about 12/14 Ã 4.05 â $3.47.
Problem 8. The prime factorization of 84 is 2
Ã 3 Ã 7, the prime factorization of 112 is 2
Ã 7, and the prime factorization of 210 is 2 Ã 3 Ã 5 Ã 7. The
greatest common factor of the three numbers is the product of all the prime factors that they have in common, which is 2 Ã 7 = 14.
Problem 9. If 10 men take 6 days to lay 1000 bricks, then 20 men should take 3 days to lay 1000 bricks. If 20 men are to lay 5000 bricks, then it should
take 5 Ã 3 = 15 days.
Problem 10. Ayushi must have 1 quarter and 5 pennies. If two coins are selected at random, there is a 5/6 Ã 1/5 = 1/6 chance that Ayushi will select a
penny and then the quarter. There is a 1/6 Ã 5/5 = 1/6 chance that Ayushi will select the quarter and then a penny. There is a 5/6 Ã 4/5 = 2/3 chance that
Ayushi will select two pennies. Only the last of these options amounts to less than 15 cents, so the probability is 2/3.
Transformation & Coordinate Geometry Stretch
Problem 1. Since only the y portions of the coordinates move, we know that the line of reîection must be a horizontal line. Now we just need to î¿nd the
midpoint between an original point and its reîected image to pinpoint the location of the line. The y-coordinate of point A is 3 and the y-coordinate of Aâ
is â5; therefore, the midpoint is at (2, â1). The line of reîection is y = â1.
Problem 2. We know that, for a triangle, area = 1/2(base)(height), which equals 30 in this problem. We also know that the height of the triangle is 4 if
we use the horizontal leg on the x-axis as the base. Now we can plug this information into the equation to î¿nd the length of the base that runs along the
x-axis. The equation is (1/2)(b)(4) = 30, so b = 30/2 = 15. Since the 3rd vertex is on the x-axis we know that it extends straight left 15 units from the
vertex at (0, 0), bringing us to the point (â15, 0).
Problem 3. When an image is translated to the right we just add the number of units it is being translated to the original x-coordinate. When an image
is translated down we just subtract that number of units from the y-coordinate. In this case weâll subtract 2 from the y-coordinates and add 3 to the
x-coordinates. This will make point B(6, 5) move to Bâ(6 + 3, 5 â 2) = (9, 3).
Problem 4. In a regular pentagon each vertex is 360Âº/5 = 72Âº away from the adjacent vertices. This means that if a point is rotated counterclockwise
144Âº, it rotates 144/72 = 2 vertices counterclockwise. Thus, vertex C would land where vertex N is.
Problem 5. When an image is reîected over an axis, the opposite coordinate changes sign. So if you translate over the y-axis, the x-coordinate changes
sign; and if you translate over the x-axis, the y-coordinate changes sign. In this case, we reîect over both the y-axis and x-axis, causing both signs to
change. Point A was originally (â10, 2), which means the î¿nal image has A at (10, â2).
Problem 6. When we rotate images 90Âº the coordinates switch places, and the signs are adjusted based on whether or not an axis was crossed. In this
case, rotating point A 90Âº will bring it across the y-axis into Quadrant I, which means both the x and y will be positive. The original point A was at (â4, 1)
so the î¿nal image will be at (1, 4). We also could solve this problem by seeing that the slope of the segment from the origin to A is â1/4. If A is moving
to a location that is a 90Âº rotation about the origin, it will move to a point on the segment perpendicular to the one that currently connects it to the origin.
This will be the segment that has a slope of 4/1 or â4/â1 from the origin which puts us at (1, 4) or (â1, â4). The point (1, 4) is in the counterclockwise
direction we need.
Problem 7. By looking at the diagram provided, we can see that the line containing the point of rotation lands on top of itself, but the arrow is facing the
opposite direction. This tells us that 1/2 of a full 360Âº rotation was completed; therefore, the image rotated 360Âº/2 = 180Âº about point C.
Problem 8. Since the image is reîected across the y-axis î¿rst, we will just change the sign of the x-coordinate, which will give us (2, 6). Next the image
is shifted down 8 units so we will subtract 8 from the y-coordinate, giving our image a î¿nal center of (2, â2).
Problem 9. With the center of dilation at the origin and a scale factor of 2, all the coordinates of square ABCD are twice the coordinates of its preimage.
The preimage has an area of 4 square units, so its side length is 2 units. Since the center of the preimage is at (8, â8), the four vertices of the preimage
are at (7, â9), (7, â7), (9, â7) and (9, â9). The point (9, â9) is the farthest from the origin on the preimage, so the point farthest from the origin on the
image of square ABCD is (18, â18).
Problem 10. Lines that are parallel have the same slope. In this case, AB has a slope of (0 â (â4))/(â4 â 0) = â1. This now must be the slope for XY.
Now we can use the equation y
) to î¿nd the value of k. Plugging in the coordinates for Y and X we î¿nd that k â 8 = â1(14 â 0), thus k =
â14 + 8 = â6. We also could see that from (0, 8) to (14, k) we are moving 14 units right, so we also must move 14 units down to get a slope of â14/14 =
â1. Moving 14 units down from (0, 8) lands us at (0, 8 â 14) or (0, â6), so k = â6.