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MATHEMATICS 320
HOMEWORK SOLUTIONS
ASSIGNMENT 2
INSTRUCTOR: MICHAEL B. SCOTT
Solution for section 12.3, B9:
The key part is figuring out what x is. First consider the transversal t passing through parallel lines l
and n. We see that the angles measuring 2x − 18◦ and x + 12◦ are supplementary since they are same side
interior angles. We didn’t explicitly derive this result about same side interior angles but you are expected
to discover this for yourself (compare ∠3 with the angles measuring 2x − 18◦ and x + 12◦ , also see problem
B11 of this section). Being supplementary means that 2x − 18◦ + x + 12◦ = 180◦ and solving we obtain
x = 62.
Now that we know x = 62 we can starting solving for the measures of ∠1, ∠2, ∠3, and ∠4. The measure
of ∠2 is x − 15◦ since the angles are vertical. This means m(∠2) = 47◦ . Notice that ∠1 and ∠2 are alternate
exterior angles, which implies they are congruent since lkm, hence m(∠1) = 47◦
Now ∠4 and ∠2 are supplementary, which implies that m(∠4) = 133◦ . Lastly, ∠3 is supplementary to
the angle measuring x + 12◦ = 74◦ , hence m(∠3) = 106◦ .
Solution for section 12.3, B10:
Part a.
Since lkm and ∠3 and ∠4 are alternate interior angles, we can say that ∠3 and ∠4 are congruent.
Furthermore, ∠1 and ∠3 are congruent because they are vertical angles. Therefore, m(∠1) = m(∠3) =
m(∠4).
Part b.
∠1 and ∠3 are congruent since they are vertical angles and ∠4 and ∠6 are congruent for the same reason.
Then, given m(∠3) = m(∠6), we can say
m(∠1) = m(∠3) = m(∠6) = m(∠4).
This means that alternate interior angles are equal, therefore lkm.
Solution for section 12.3, B11:
Part a.
We can conclude that m(∠1) + m(∠3) = 180◦ . We know that m(∠1) + m(∠2) = 180◦ , because ∠1
and ∠2 add up to a straight line. Since lkm and ∠2 and ∠3 are alternate interior angles, we have that
m(∠2) = m(∠3). Now we can say m(∠1) + m(∠3) = 180◦ .
Part b.
The converse of part (a) is true. That is, if same side interior angles are supplementary, then lkm. To see
this we subtract the following equations:
m(∠1) + m(∠3) = 180◦
− m(∠1) + m(∠2) = 180◦
Which is given.
Since∠1 and ∠2 form a straight line.
0 + m(∠3) − m(∠2) = 0
Thus m(∠3) = m(∠2) which means that alternate interior angles are equal. Therefore, lkm.
Solution for section 13.1, B2cfh:
Part c.
A square mile is a region one mile by one mile and an acre is a plot of land with an area of 43,560 square
feet. One mile is 5280 feet and hence a square mile has area 5280 × 5280 = 27, 878, 400 square feet. The
Date: Spring 2004.
1
2
INSTRUCTOR: MICHAEL B. SCOTT
number of acres in a square mile is thus the number of 43,560 square foot plots it will take to cover this area
of 27,878,400 square feet. Using unit analysis:
1 mile2 ·
27, 878, 400 ft2
1 acre
·
= 640 acres.
2
1mile
43, 560 ft2
Part f.
There are 2 cups in 1 pint, so a half-pint is 1 cup.
Part h.
There are many ways to do this problem. We will ounces as our base measurements. There are 2
tablespoons in 1 fluid ounce, 3 teaspoons in 1 tablespoon, and 8 fluid ounces in 1 cup. Using unit analysis
we have
8 fl. oz. 2 Tbs. 3 Tsp.
1 cup ·
·
·
= 48 Tsp.
1cup
1 fl. oz. 1 Tbs.
Solution for section 13.1, B3ae:
Remark. The are two steps when converting between Celsius and Fahrenheit.
First, there are 1.8◦ F for each Celsius degree. One way to see this is that the freezing and boiling points
of water for Celsius are 0◦ C and 100◦ C respectively and for Fahrenheit are 32◦ F and 212◦ F. This means
there are 100 units of Celsius for 180 units of Fahrenheit.
Second the zeros of each temperature unit are different. Just remember that 0◦ C and 32◦ F both correspond
to the freezing point of water.
Part a.
In this problem we have to convert 136◦ F to Celsius. Since the difference in the zeros of the temperature
units are off by 32◦ F we always subtract or add 32◦ F to the Fahrenheit units to deal the zeros being off. In
this case we have to subtract 32◦ F from 136◦ F which is 104◦ . Now, using unit analysis, we convert the units
1◦ C
= 57.8◦ C or 58◦ C.
104◦ F ·
1.8◦ F
Part e.
In this problem we have to convert −18◦ C to Fahrenheit, so we have to convert the unit size first. That
is,
1.8◦ F
−18◦ C · ◦
= −32.4◦ F.
1 C
Now, we “adjust the zero” of Fahrenheit by adding 32◦ F to −32.4◦ giving a answer of −0.4◦ F.
Solution for section 13.1, B15a:
The first thing we must do is find the volume of the prism, which is 24 × 18 × 9 = 3, 888 cubic inches.
Now, we can use unit analysis to compute how much the water weighs, that is,
1 gal. 8.3 lbs.
3, 888 in.3 ·
·
= 139.7 lbs..
231 in.3 1 gal.
Solution for section 13.1, B17:
Part a.
The student has used an inappropriate ratio by using 1 ft./12 in. It should read
12 in.
= 22 · 12 in. = 264 in.
22 ft. = 22 ft. ·
1 ft.
Part b.
Treat the units as fraction. The unwanted units should cancel out leaving just the desired units. Notice
that in this problem the feet units cancel and only the inch unit remains.
Solution for section 13.2, B6abc:
If you draw lines between each of the lattice points you will get a set of squares and triangles. Each square
counts as 1 unit and each triangle counts as a 1/2 unit. In part (a) you should get an area of 25, part (b)
an area of 20, and in part (c) an area of 22 12 .
Solution for section 13.2, B7:
First let the distance between any two horizontal or vertical lines between lattice points be length 1.
Then, in this problem you have to recognize that a 1 × 2 rectangle is double the size of the given unit area.
To see this draw a 1 × 2 rectangle and you can draw two unit area triangles inside. This means a 1 × 1 square
is equal to the given unit area and that a triangle with base 1 and height 1 is 1/2 the given unit area.
MATHEMATICS 320
HOMEWORK SOLUTIONS
ASSIGNMENT 2
3
Using the ideas above we can count the area of the given shapes. In part (a) the area is 7, part (b) the
area is 6 12 , and in part (c) the area is 8.
Solution for section 13.2, B16ab:
Part a.
This is only one of many ways to do this problem. The area of the trapezoid is 2. It helps to know that
the area of the 2 × 1 triangle is equal to 1 square unit.
Part b.
This is only one of many ways to do this problem. Recognize that the original trapezoid can be partitioned
by a unit square and a 2 × 1 triangle is equal to 1 unit square. This means the area of the trapezoid is 2
square units. Now, if we triple the dimensions of a square unit to a 3 × 3 square we have an area of 9. Since
the trapezoid has an area of 2 square units, its area when its dimensions are tripled must be 2 · 9 = 18.
Solution for section 13.2, B17b:
√
This problem is an application of the Pythagorean theorem. To represent 17 on a square lattice you need
draw a right triangle with the appropriately sized legs. Recall that the Pythagorean theorem characterizes
the relationship between the sides of a right triangle via the equation a2 + b2 = c2 where a and b are the legs
of the triangle and c is the hypotenuse.√
Suppose we have triangle with c = 17, then c2 = 17. Can we find two numbers when squared that add
up to 17? Consider the squares of the counting numbers less than 17, that is, 1, 4, 9, and 16. Notice that
1 + 16 = 17. This means if we draw a√right triangle on a square lattice with legs 1 and 4, then the line of
the hypotenuse represents a length of 17.
Department of Mathematics, Kansas State University
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