Download Math 202: Homework 4 Solutions

Math 202: Homework 4 Solutions
1. (§10.2 #8)
(a) Convex
(b) Concave
(c) Convex
(d) Concave
2. (§10.2 #34) We are told that the trapezoids are isosceles, so both of the nonparallel
sides of the first trapezoid are 6 units long, and both of the nonparallel sides of the
second trapezoid are x units long. The ratio of the lengths of each of the longer
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18
= , and the ratio of the lengths of each of the nonparallel sides
parallel sides is
12
2
6
is . We are told that these ratios are equal, so
x
3
6
=
2
x
6
3
·x = ·x
2
x
3
x=6
2
2
2 3
· x = ·6
3 2
3
x=4
3. (§10.3 #4)
(a) This figure is not a polygon, because it is not closed.
(b) This figure is not a polygon, because one of its sides is not straight.
(c) This figure is a polygon, because it is a closed figure with all straight sides.
4. (§10.3 #8)
(a) This figure is a 6-gon, i.e. a hexagon, so the sum of its interior angle measures
is
180◦ (6 − 2) = 180◦ (4) = 720◦
(b) This figure is an 8-gon, i.e. an octagon, so the sum of its interior angle measures
is
180◦ (8 − 2) = 180◦ (6) = 1080◦
5. (§10.3 #10)
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(a) This polygon is regular, because the markings on it indicate that all its sides are
congruent, and all its interior angles are congruent.
(b) This polygon is regular, because the markings on it indicate that all its sides are
congruent, and all its interior angles are congruent.
(c) This polygon is not regular, because if one of its angles is 58◦ , and the others are
congruent to each other, then the others are each 61◦ . Hence, not all the interior
angles are congruent.
(d) This polygon is an octagon, and we are told that one of its interior angle measures is 140◦ . Since the sum of its interior angle measures is 180◦ (8 − 2) =
1080◦ , then each interior angle measure in a regular octagon should be 1080◦ ÷
8 = 135◦ . Since our one known angle is not 135◦ , we know this cannot be a
regular octagon.
6. (§10.3 #12)
(a) The sum of the interior angle measures of a hexagon is 180◦ (6 − 2) = 720◦ , so
each of the interior angles in a regular hexagon must measure 720◦ ÷ 6 = 120◦ .
Similarly, each central angle must measure 360◦ ÷ 6 = 60◦ .
(b) The sum of the inferior angle measures of a heptagon is 180◦ (7 − 2) = 900◦ ,
so each of the interior angles in a regular heptagon must measure 900◦ ÷ 7 ≈
128.6◦ . Similarly, each central angle must measure 360◦ ÷ 7 ≈ 51.4◦ .
7. (§10.3 #14)
(a) This polygon is a heptagon, so the sum of its interior angle measures must be
180◦ (7 − 2) = 900◦ . Hence,
112 + 123 + 147 + 127 + 127 + x + 126 = 900
x + 762 = 900
x = 900 − 762
x = 138
(b) This polygon is a hexagon, so the sum of its interior angle measures must be
180◦ (6 − 2) = 720◦ . Hence,
x + 144 + 141 + 133 + 90 + 90 = 720
x + 598 = 720
x = 720 − 598
x = 122
8. (§10.3 #22) It seems that the student mixed up the formulas for each interior angle
measure and for each central angle measure of a regular polygon. The formula
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to find each central angle measure should not include an n in the denominator. I
would encourage the student to review the intuition behind these angle formulas,
so he/she could catch a mistake like this in the future.
9. (§10.3 #26) Note that, for every angle in a quadrilateral to be congruent, this means
that each angle is 90◦ . Hence, to get a polygon with congruent angles that are obtuse,
we’ll need at least five sides. If we want the sides not to be congruent, we should
start by drawing a regular pentagon, then stretch it a bit, while maintaining all the
angles, like so:
hi
10. (§10.3 #28) We need only notice that each of the angles in each triangle at least contributes to an interior angle, and each interior angle is made up completely of the
angles in these triangles. Hence, the sum of the angle measures in all the triangles
should give us the sum of the interior angle measures of the polygon. Since the sum
of the angle measures in each triangle is 180◦ , and we see in this case that there are
4 triangles, we know that the sum of the interior angle measures of the hexagon is
180◦ (4) = 720◦ .
Looking at a few more examples, one will notice that adding a side to the polygon
also adds a triangle to its inside, thus increasing the interior angle measure sum by
180◦ (and justifying the 180◦ (n − 2) formula).
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11. (§10.3 #20)
12. (§10.3 #34) No. Each interior angle in a hexagon is 120◦ , which divides 360◦ into
three. Since each interior angle measure becomes larger as the number of sides
increases, the next possible angle measure that could divide 360◦ is 180◦ . However,
if each interior angle were 180◦ , then the polygon would be one straight line, which
could never be closed. Hence, there is no such polygon.
13. (§10.4 #4)
(a) Yes. It seems from the picture that every face of the solid is either a square or a
triangle.
(b) No. This solid involves curves, so its faces are not polygons.
(c) Yes. This seems to be a quadrilateral pyramid.
(d) No. This solid does not have polygons as faces.
14. (§10.4 #6)
(a) This solid is not a polyhedron, because it has curves, so it is certainly neither a
prism nor a pyramid.
(b) This solid is a pyramid. Its base is a (concave) 12-gon.
(c) This solid is a prism with rectangular base.
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(d) This solid is a pyramid with hexagonal base.
15. (§10.4 #8)
(a) This is an octagonal prism.
(b) This is a decagonal prism.
16. (§10.4 #10)
This polyhedron has 8 vertices, 12 edges, and 6 faces.
17. (§10.4 #12)
(a) First, notice that we are given the number of lateral faces, and since this is a
prism, we know there are 12 faces total. Now we use Euler’s Formula:
V+F=
20 + 12 =
32 =
32 − 2 =
30 =
E+2
E+2
E+2
E
E
So a decagonal prism has 30 edges.
(b) Notice again that we are given the number of lateral faces, and since this is a
pyramid, we know there are 5 faces total. Now we use Euler’s Formula:
V+F = E+2
V+5 = 8+2
V + 5 = 10
V = 10 − 5
V=5
So a pyramid with a kite base has 5 vertices.
(c) We use Euler’s Formula:
V+F
10 + F
10 + F
F
F
5
= E+2
= 18 + 2
= 20
= 20 − 10
= 10
So this polyhedron has 10 faces.
18. (§10.4 #20) The ramp looks like a triangular prism. If so, it has 6 vertices, 9 edges,
and 5 faces.
19. (§10.4 #26) The student got Euler’s Formula mixed up, saying that F + E = V + 2,
when, in fact, Euler’s Formula really states that V + F = E + 2. I would encourage
the student to review Euler’s Formula and test it on small, easy examples if he ever
forgets.
20. (§10.4 #28)
(a) We know the prism has 2 bases and one lateral face for each side of the base.
Since the base has n sides, then there must be n + 2 faces total.
(b) There are n vertices in each base and no other vertices in between. Hence, there
are 2n vertices total.
(c) There are n edges in each base, and there are n additional edges on the lateral
faces. Hence, there are 3n edges total.
(d)
V+F = E+2
2n + (n + 2) = 3n + 2
3n + 2 = 3n + 2
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