Test 2 Key
1 Dilations can change distance. Choice 1
2 A cylinder could model the shaft of the pencil and a cone could be used to model the tip. Choice 4
3 Choice 3 if the right triangle below is rotated around the vertical leg, the result is a cone
4 Choice 2 A diagonal connects nonconsecutive vertices. A side connects consecutive vertices and the
other two choices are not line segments at all
5 The area of the floor = Area of the rectangle – Area of the two circles
= length x width – 2(𝜋𝑟 2 )
= Choice 4
6 Choice 3 because 3(2√10) = 6√10
7 Sin and Cos values are = when the angles are complimentary (90). So:
3x + 2 + 4x – 10 = 90
7x -8 = 90
7x = 98
X = 14.0, choice 3
8 Choice 1. A dilation will result in a similar figure which has equal angle measures but different size.
9 <BCD is an inscribed angle to a semicircle so that = 90
<BED is an inscribed angle to a semicircle so that = 90
<BDE is an inscribed angle but not to a semicircle so it is not = 90
<FBD is formed by a tangent and a diameter and is = .5(180) = 90
Choice 3
10 Slope =
3−8
7−2
=
4
−5
5
4
5
and a 4:1 ratio means we need to get of the way from point 1 to point 2 so,
4
X = 𝑥1 + 5(5) = 2 +5(5) = 6
4
4
Y = 𝑦1 + 5(-5) = 8 + 5(-5) = 4
Therefore, (6, 4) is the point that divides the segment into a 4:1 ratio. Choice 3
1
11 Perpendicular slopes are negative reciprocals of each other. The slope of the given line y = x – 2 is 1 .
1
The negative reciprocal of 1 is
−1
1
. Choice 2 is the only answer that has a slope = -1.
12 Based on the given equation the center is at (3, -1) and the radius = √16 = 4. Choice 1 is the graph of
the circle with the center at (3, -1) and r = 4.
13 Since 1 oz = 1.8 cu. in, 40 oz = 40(1.8) = 72 cu. in. and r = .5(3) = 1.5
So, 72 = 𝜋𝑟 2 h
72 = 𝜋1.52 h
72
(𝜋1.52 )
=h
10.2 = h
Choice 3
14 The center of the rectangle is 5.5” from the left edge and 7” from the bottom edge.
Because of the golden ratio 1:1.618, the distance from the left edge to B is
and the distance from the bottom edge to B is
1
(14)
1+1.618
1
(11) =
1+1.618
4.201680672
= 5.347593583. If we think of the lower left
hand corner of the rectangle as the origin (0, 0), then B = (4.201680672, 5.347593583) and the center of
the rectangle = (5.5, 7). We need to calculate the distance between these two points using the distance
formula D = √((5.5 − 4.201680672)2 + (7 − 5.347593583)2 ≈ 2.1. Choice 3.
15 If a cylinder stands on one of its circular bases. The horizontal cross-section is a circle, the vertical
cross-section is a rectangle, and an angled cross section is an ellipse. Choice 4 is the only one that is not
possible.
16 Choice 1. As long as all cross sections have equal areas and volume is calculated by multiplying that
sum by the height, the volumes are therefore equal as well.
17 The more sides that a figure has, the fewer degrees of rotational symmetry it has. The regular
octagon has 8 sides. That is more than any other choice and therefore has the smallest number of
degrees in its rotational symmetry.
18 Choice 3.
19 Put each in y= form and compare :
3y – x = 3
y + 3 = 3(x – 1)
3y = x + 3
y + 3 = 3x - 3
1
Y = 3x + 1 and
y = 3x -6
1
Slope= 3
Slope = 3
choice 4 (not = so not parallel and not negative reciprocals so not perp.)
20 Choice 2
21 Graph each set of coordinates. For the rectangle to have an area = 50, we need to know the length of
AB = √(5 − 1)2 + (−1 − 2)2 = 5. So CD must be 10 and must be perpendicular to AB. Choice 2 is both.
22 Start with ABC and do each set of transformations and you will realize it is Choice 3.
23 Choice 1.
24 Choice 2.
25 CA is a radius with length = √(6 − 0)2 + (2 − 5)2 = √45 = 3√5. So if B is on the circle CB must = 3√5
CB = √(11 − 5)2 + (−3 − 0)2 = √45 = 3√5. Therefore, B is on the circle.
26 Construct 60 degree angles. One will pass through T. See construction.
27 A =
𝜋(8)2
8
vs.
𝜋(8)2
12
= 25.133 - 16.755 = 8.378≈ 8.4sq. in.
28 See construction.
29 See Diagram
Statement
1. Isosceles ∆ABC, bisector AD
2. AB ≅ AC
3. AD ≅ AD
4. BD ≅ DC
5. ∆𝐴𝐷𝐵 ≅ ∆ADC
6. <B ≅ <C
30 2x + 10 = 5x – 5
10 = 3x – 5 so 15 = 3x and x = 5.
Reasons
Given
Def. of Isosceles ∆
Reflexive
Def. of Perp. bisector
SSS
CPCTC
31 Equidistant from 3 points is the circumcenter. Construct the perpendicular bisectors. See
construction.
1
32 Compare the formulas, the volume of the cone is 3 that of the cylinder.
33 Graph the points. Prove the quadrilateral is a parallelogram by showing both pairs of opposite sides
parallel using the slope formula. Then you can show that it is a rectangle by noting that the consectutive
slopes are negative reciprocals and therefore perpendicular, thus forming right angles.
3− −2
5
Slope AB = −2− −6 = 4
Slope CD =
−6− −1 5
=4
−1−3
−1−3
Slope BC = 3− −2 =
−6− −2
−4
5
Slope AD = −1− −6 =
−4
5
a) AB ll CD and BC ll AD because slopes are = respectively. Therefore, ABCD is a parallelogram.
b) AB is perpendicular to BC because their slopes are negative reciprocals. Therefore, <B is a right
angle and ABCD is a rectangle.
34 Construct the perpendicular bisector of each side and connect the midpoint of each to the vertex on
the opposite side. See construction. There is only 1 point of intersection.
35 Construct the perpendicular bisector to the diameter. Draw the square so that the diagonals are
perpendicular by construction and congruent because they are both diameters.
The area of the square can be found by calculating the area of the 4 triangles formed within the square:
A = .5(bh)
.5(12.5)(12.5)
78.125
4 X 78.125 = 312.5 SQ IN. TOTAL for the 4 triangles within the square
36a. To get from R to R’ you must dilate by a scale factor of 2. The constant of dilation is 2.
b. The constant of dilation gives you the similarity ratio used to show the triangles are similar.
37 See diagram
a.
b.
c.
d.
Translation (x – 3, y + 2)
Reflection in the y-axis
Rotation of 90 degrees about the origin
Yes, each of the transformations is a rigid motion and therefore you can get from ABC to
A’’’B’’’C’’’ using a sequence of rigid motions
e. A dilation of any kind, say a dilation by 2 would result in a triangle that is not congruent to the
others.