Checkpoint week 4
Math/157
Section 10.1 Problems: 10,12,20,26,32
Problem: 10
These are opposite a given vertex.
Angles DHG and FHM
Problem 12
An obtuse angle is larger than 90 degrees:
GHM
Problem 32
No, there isn’t only one plane. You can imagine a plane through the line getting rotated
around, using the line as an axis. There are infinitely many of these planes.
Problems: 20
It is true because all the points on a line always lie in the same plane.
Problem 26
True, because a right angle is 90 degrees, leaving only 90 degrees between the other two
angles, and an obtuse angle has to be larger than 90 degrees.
Section 10.2 Problems: 2,8,34
Problems: 2
The centroid of a triangle is where the three medians meet. It’s also the point on the
triangle that you can use to balance it, since it’s the center of mass if the triangle was uniform
density.
Problem 8
The ratio of the portions of the median on either side of the centroid is 2:1.
Therefore, the centroid lies 24 units from one end of the median, and 12 units from the
other.
Problem 34
Since there are many “odd” vertices, it is impossible to walk all the bridges and return
home without repeating yourself.
Section 10.3 Problems: 2,4,22,26,28,32,36,48
Problems: 2
Reasons:
A: vertical angles
B: straight lines add to 180 deg
C: straight lines add to 180 deg
D: alternate interior angles
E: using angle 5 and straight lines adding to 180 degrees
F: vertical angles from angle 5
G: vertical angles from angle 6
Answers:
A: 52 degrees
B: 128 degrees
C: 128 degrees
D: 52 degrees
E: 128 degrees
F: 52 degrees
G:128 degrees
Problem 4
Angles 1 and 7 are congruent, by alternate exterior angles.
4x+10=6x+4
10 = 2x + 4
6 = 2x
X=3
The angles are:
4*3 + 10 = 12+10 = 22
Therefore, they are 22 degrees each
Problems: 22
The central and exterior angles are the same.
They are both 360/n.
The interior is 180(n-2)/n
Taking a ratio of the central to the interior:
(360/n)/(180(n-2)/n)
= (360/(180(n-2))
= 2/(n-2)
Therefore, the ratio of the central to the interior angle is 2/(n-2)
Problem 26
The sum of the angles in a quadrilateral is:
180(n-2) = 180(4-2) = 180*2 = 360 degrees
360-4*87
= 12 deg
Problem 32
The formula is:
P = ½(m(arc a) - m(arc b))
= ½*(210-150)
= ½*(60)
= 30 deg
Problem: 28
By definition, a triangle formed from the diameter of a circle, and two other lines that hit the circle
at a point make a right triangle. There are two of these types of right triangles in this diagram.
Problem: 36
The arc length on both sides of the lighthouses is 180 degrees. If the difference between the
angles is 90 degrees, he is at the danger area (half of 180). Therefore, the difference has to be less than
90 degrees.
Problem: 48
Use the exterior angle formula:
40 = 360/n
40n = 360
N = 360/40
N=9
Section 10.4 problems: 2,10,32,44,72
Problem 2
AHJ is congruent to GOP
By the SSS theorem
Problem 10
MKI is congruent to MLI
By the SSS theorem
Problem: 32
The sides for a 30-60-90 triangle are:
1-sqrt(3)-2
Since the shorter leg is 10, we have to multiply by 10:
10-10sqrt(3)-20
Using a calculator for the sqrt(3):
10-17.3-20
So the longer leg is 17.3 and the hypotenuse is 20.
Problem: 44
Reasons:
A: the rule is 1-1-sqrt(2)
B: the rule is 1-sqrt(3)-2
C: as with part B
D: the is the inverse of the previous part
E: THERE IS A TYPO IN THE TEXTBOOK‼! The question is asking you to solve for itself…
Answers:
A: √2
B: 2
C: √3
D: 1/√3
E: The question doesn’t make sense, because there is definitely a typo in the textbook
Problem: 72
Reasons:
A: the side c gets shorter, and “a” and “b” grow
B: the side c gets longer, and “a” and “b” shrink
Answers:
A: a^2+b^2 is now more than c^2
B: a^2+b^2 is now less than c^2
Section 10.5 Problems: 2,6,26,44
Problems: 2
For a rectangle:
Angles are equal
Opposite sides are equal
Diagonals are equal
All the angles are 90 degrees
AB = CD
AC = BD
AD = BC
AE = ED = BE = CE
Problem 6
There are three types that have perpendicular bisectors: Kite, rhombus, square
Problem: 26
In a parallelogram, the diagonals always bisect each other. Therefore, the floor and ironing
surface are parallel to each other.
Problem: 44
Since it’s a parallelogram, the opposite sides are already parallel. If they are also congruent (as
the problem states), then we have the definition of a rhombus.