Download 155 6–2 Triangles Interior Angles of a Polygon

Section 6–2
◆
155
Triangles
2.16
2.01
B
Parallel
x
5.43
A
90°
35°
Line
FIGURE 6–15
FIGURE 6–14
ight
of s
Angle of elevation
Horizontal
Observer
Ra
y
(a) Angle of elevation
D
A
x
C
Horizontal
5
2.5
B
m
Observer
2.25 m
46.3°
Line
Angle of depression
of si
ght
4.06 m
(b) Angle of depression
FIGURE 6–16
6–2
FIGURE 6–17
FIGURE 6–18
Triangles
Polygons
A polygon is a plane figure formed by three or more line segments, called the sides of the
polygon, joined at their endpoints, as in Fig. 6–19. The points where the sides meet are called
vertices. We say that the sides of a polygon are equal if their measures (lengths) are equal. If all
of the sides and angles of a polygon are equal, it is called a regular polygon, as in Fig. 6–20.
The perimeter of a polygon is simply the sum of its sides.
Vertex
A
Side
Interior
angle
FIGURE 6–19
B
A polygon
Equilateral triangle
Square
Pentagon
Hexagon
FIGURE 6–20 Some regular polygons.
Sum of Interior Angles
A polygon of n sides has n interior angles, such as those shown in Fig. 6–21. Their sum is equal
to the following:
Interior Angles
of a Polygon
Sum of angles (n 2)180
112
Modern definitions of plane
figures exclude the interior as
part of the figure. Thus in
Fig. 6–20, point A is not on the
square, while point B is on the
square. The interior is referred to
as a region.
156
Chapter 6
123˚
◆
Geometry
◆◆◆
99˚
Example 4: Find angle in Fig. 6–21.
Solution: The polygon shown in Fig. 6–21 has seven sides, so n 7. By Eq. 112,
Sum of angles (7 2)180 900
62˚
278˚
Adding the six given angles gives us
226˚
278 62 123 99 226 43 831
43˚
So
900 831 69
FIGURE 6–21
Scalene triangle
Acute triangle
Isosceles triangle
◆◆◆
Triangles
A triangle is a polygon having three sides. The angles between the sides are the
interior angles of the triangle, usually referred to as simply the angles of the triangle.
As shown in Fig. 6–22, a scalene triangle has no equal sides; an isosceles triangle
has two equal sides; and an equilateral triangle has three equal sides.
An acute triangle has three acute angles; an obtuse triangle has one obtuse
angle and two acute angles; and a right triangle has one right angle and two acute
angles.
Obtuse triangle
Altitude and Base
The altitude of a triangle is the perpendicular distance from a vertex to the opposite
side called the base, or an extension of that side (Fig. 6–23).
Equilateral triangle
Right triangle
FIGURE 6–22. Types of triangles.
The measuring of areas,
volumes, and lengths is
sometimes referred to as
mensuration.
Area of a
Triangle
Altitude
h
Altitude
h
Base b
Area of a Triangle
We use the following familiar formula:
Base b
Area equals one-half the product of the base and the
altitude to that base
bh
A 2
137
◆◆◆
Example 5: Find the area of the triangle in Fig. 6–23 if the base is 52.0
and the altitude is 48.0.
Solution: By Eq. 137,
FIGURE 6–23
52.0(48.0)
area 1250
2
◆◆◆
(rounded)
If the altitude is not known but we have instead the lengths of the three sides, we
may use Hero’s formula. If a, b, and c are the lengths of the sides, the following formula
applies:
This formula is named for Hero
(or Heron) of Alexandria, a Greek
mathematician and physicist of
the 1st century A.D.
Hero’s
Formula
Area of triangle s(s a)(s b)(s c)
where s is half the perimeter, or
abc
s 2
138
Section 6–2
◆◆◆
◆
157
Triangles
Example 6: Find the area of a triangle having sides of lengths 3.25, 2.16, and 5.09.
Solution: We first find s, which is half the perimeter.
3.25 2.16 5.09
s 5.25
2
Thus the area is
area 5.25(5.25 3.25)(5.25 2.16)(5.25 5.09) 2.28
◆◆◆
Sum of the Angles
An extremely useful relationship exists between the interior angles of any triangle.
The sum of the three interior angles A, B, and C of any
triangle is 180 degrees.
Sum of the
Interior
Angles
◆◆◆
139
A B C 180
Example 7: Find angle A in a triangle if the other two interior angles are 38 and 121.
Solution: By Eq. 139,
A 180 121 38 21
◆◆◆
Exterior Angles
An exterior angle is the angle between the side of a triangle and an extension of the adjacent side, such as angle in Fig. 6–24. The following theorem applies to exterior angles:
The exterior angle of a triangle is the sum of the two
angles in the triangle that are not the angle along the
extension of the adjacent side.
Exterior
Angle of a
Triangle
B
=A+B
A
C
FIGURE 6–24 An exterior
angle.
142
AB
Congruent Triangles
Two triangles (or any other polygons, for that matter) are said to be congruent if the angles and
sides of one are equal to the angles and sides of the other, as in Fig. 6–25.
2
4
5
5
2
4
FIGURE 6–25 Congruent triangles.
Similar Triangles
Two triangles are said to be similar if they have the same shape, even if one triangle is
larger than the other. This means that the angles of one of the triangles must equal the
angles of the other triangle, as in Fig. 6–26. Sides that lie between the same pair of equal
angles are called corresponding sides, such as sides a and d. Sides b and e, as well as
sides c and f, are also corresponding sides. Thus we have the following two theorems:
c
a
b
f
d
e
b
c
a
=
=
e
f
d
FIGURE 6–26 Similar
triangles.
158
Chapter 6
We will see in a later chapter
that these relationships also
hold for similar figures other than
triangles, and for similar solids
as well.
A
Geometry
Similar
Triangles
B
5.14 m
◆
143
Corresponding sides of similar triangles are in
proportion.
144
Example 8: Two beams, AB and CD, in the framework of Fig. 6–27 are parallel.
Find distance AE.
◆◆◆
Solution: By statement 104, we know that angle AEB equals angle DEC. Also, by
statement 105, angle ABE equals angle ECD. Thus triangle ABE is similar to triangle
CDE. Since AE and ED are corresponding sides,
E
5.8
AE
5.87
7m
7.25 m
C
If two angles of a triangle equal two angles of another
triangle, the triangles are similar.
5.14
7.25
5.14
AE 5.87 p q 4.16 m
7.25
D
◆◆◆
FIGURE 6–27
Right Triangles
In a right triangle (Fig. 6–28), the side opposite the right angle is called the hypotenuse, and
the other two sides are called legs. The legs and the hypotenuse are related by the well-known
Pythagorean theorem.
Pythagorean
Theorem
B
c
A
a
+
b2
145
a2 b2 c2
C
b
a2
The square of the hypotenuse of a right triangle is
equal to the sum of the squares of the two legs.
=
c2
◆◆◆
FIGURE 6–28. The
Pythagorean theorem, named
for the Greek mathematician
Pythagoras (ca. 580–500 B.C.).
Example 9: A right triangle has legs of length 6 units and 11 units. Find the length of the
hypotenuse.
Solution: Letting c the length of the hypotenuse, we have
c2 62 112 36 121 157
c 157 12.5 (rounded)
◆◆◆
◆◆◆
Example 10: A right triangle has a hypotenuse of 5 units in length and a leg of 3 units.
Find the length of the other leg.
Solution:
a2 c2 b2
52 32 16
a 16 4
Common
Error
Remember that the Pythagorean theorem applies only to
right triangles. Later we will use trigonometry to find the
sides and angles of oblique triangles (triangles with no right
angles).
◆◆◆
Section 6–2
◆
159
Triangles
Some Special Triangles
In a 30–60–90 right triangle [Fig. 6–29(a)], the side opposite the 30 angle is half the length of
the hypotenuse.
A 45° right triangle [Fig. 6–29(b)] is also isosceles, and the hypotenuse is 2 times the
length of either side.
A 3–4–5 triangle [Fig. 6–29(c)], is a right triangle in which the sides are in the ratio of
3 to 4 to 5.
b=
c
2
60˚
x
1
6
b
45˚
30˚
a
(a)
1
(b)
FIGURE 6–29
◆◆◆
45˚
公2
c
A
8
(c)
Some special right triangles.
Example 11: In the 3–4–5 triangle of Fig. 6–29(c), side x is 10 cm.
Exercise 2
◆
◆◆◆
Triangles
1. What is the cost, to the nearest dollar, of a triangular piece of land whose base is 828 m and
altitude 412 m, at $2,925 per acre? Assume all the figures are exact.
2. The house shown in Fig. 6–30 is 12.0 m wide. The ridge is 4.50 m higher than the side
walls, and the rafters project 0.750 m beyond the sides of the house. How long are the
rafters?
3. What is the length of a handrail for a flight of stairs (Fig. 6–31) where each step is 30.5 cm
wide and 22.3 cm high?
il
dra
n
Ha
16th step
30.5 cm
22.3 cm
0.750 m
?
4.50 m
1st step
12.0 m
FIGURE 6–30
The 3–4–5 right triangle is a
good one to remember, as it is in
regular use.
FIGURE 6–31
4. At $13.50 per square metre, find to the nearest dollar the cost of paving a triangular court,
its base being 32.0 m and its altitude 21.0 m.
5. A vertical pole 15.0 m high is supported by three guy wires attached to the top and reaching
the ground at distances of 20.0 m, 36.0 m, and 66.7 m from the foot of the pole. What are
the lengths of the wires?
6. A ladder 13.0 m long reaches to the top of a building when its foot stands 5.00 m from the
building. How high is the building?
7. Two streets, one 16.2 m and the other 31.5 m wide, cross at right angles. What is the
diagonal distance between the opposite corners?
For practice, do these problems
using only geometry, even if you
know some trigonometry.
To help you solve each problem,
draw a diagram and label it
completely. Look for rectangles,
special triangles, or right
triangles contained in the
diagram. Be sure to look up any
word that may be unfamiliar,
such as diagonal, radius, or
diameter.
160
Sla
nt h
21. eight
8 in
.
Chapter 6
18.4 in.
FIGURE 6–32
We have not studied the hexagon
as such, but you can solve
problems 15 and 16 by dividing
the hexagon into right triangles
or a rectangle and right triangles.
This also works for other regular
polygons.
◆
Geometry
8. A room is 5.00 m long, 4.00 m wide, and 3.00 m high. What is the diagonal distance
from one of the lower corners to the opposite upper corner?
9. What is the side of a square whose diagonal is 50.0 m?
10. A rectangular park 125 m long and 233 m wide has a straight walk running
through it from opposite corners. What is the length of the walk?
Height
11. A ladder 8.00 m long stands flat against the side of a building. How far must it be
drawn out at the bottom to lower the top by 1.00 m?
12. The slant height of a cone (Fig. 6–32) is 21.8 in., and the diameter of the base is
18.4 in. How high is the cone?
13. A house (Fig. 6–33) that is 25.0 m long and 20.0 m wide has a pyramidal roof
whose height is 7.50 m. Find the length PQ of a hip rafter that reaches from a
corner of the building to the vertex of the roof.
14. An antenna, 125 m high, is supported by cables of length L that reach from the top
of the tower to the ground at a distance D from the base of the antenna. If D is 43
the length of a cable, find L and D.
15. A hex head bolt (Fig. 6–34) measures 2.00 cm across the flats. Find the distance x across
the corners.
16. A 1.00-m-long piece of steel hexagonal stock measuring 30.0 mm across the flats has a
10.0-mm-diameter hole running lengthwise from end to end. Find the mass of the bar
(density 7.85 g/cm3).
17. Find the distance AB between the centres of the two rollers in Fig. 6–35.
P
7.50 m
23.4 cm
diameter
Q
A
B
.0
20
2.00
cm
15.8 cm
diameter
m
0m
25.
x
FIGURE 6–33
38.8 cm
FIGURE 6–34
FIGURE 6–35
18. A 0.500-m-long hexagonal concrete shaft measuring 20.0 cm across the corners has a
triangular hole running lengthwise from end to end (Fig. 6–36). The triangular hole is in
the shape of an equilateral triangle with sides 10.0 cm long. What is the total surface area
of the shaft?
Side view
00
0.5
End view
m
10.0 cm
20.0 cm
FIGURE 6–36
Section 6–2
◆
161
Triangles
1874 ft.
C
ay
hw ft.
77
28
g
Hi
A
2025 ft.
111.8°
C
B
100.4°
A
B
FIGURE 6–37
right angle.
Note: Angle B is not a
FIGURE 6–38
19. A highway (Fig. 6–37) cuts a corner from a parcel of land. Find the number of acres in the
triangular lot ABC. (1 acre 43 560 sq. ft.)
20. A surveyor starts at A in Fig. 6–38 and lays out lines AB, BC, and CA. Find the three
interior angles of the triangle.
21. A beam AB is supported by two crossed beams (Fig. 6–39). Find distance x.
2.16 m
A
B
x
Parallel
5.16 m
FIGURE 6–39
4.25 m
22. Find dimension x in Fig. 6–40.
23. Find dimension x on the beveled end of the shaft in Fig. 6–41.
8.0 mm
59 mm
45˚
29 mm
150˚
17 mm
R
30°
92 mm
17 mm
27 mm
57 mm
35.0 mm
22 mm
x
x
FIGURE 6–40
FIGURE 6–41
162
Chapter 6
6–3
◆
Geometry
Quadrilaterals
A quadrilateral is a polygon having four sides. Quadrilaterals are the familiar figures shown in
Fig. 6–42. The formula for the area of each region is given right on the figure. Lengths of line
segments are measured in units of length (such as metres, centimetres, and inches), and areas
of regions are measured in square units (such as square metres, square centimetres, and square
inches).
b
a
a
(a) Quadrilateral
Area = a2
a
(b) Square
Area = ab
(c) Rectangle
a
Area = ah
a
h Area = bh
a
h
a
h
Area =
b
(d) Parallelogram
(e) Rhombus
(a + b) h
2
b
(f) Trapezoid
FIGURE 6–42 Quadrilaterals.
For the parallelogram, opposite sides are parallel and equal. Opposite angles are equal, and
each diagonal cuts the other diagonal into two equal parts (they bisect each other).
The rhombus is also a parallelogram, so the previous facts apply to it as well. In addition,
its diagonals bisect each other at right angles and bisect the angles of the rhombus.
The trapezoid has two parallel sides, which are called the bases, and the altitude is the
distance between the bases.
◆◆◆ Example 12: A solar collector array consists of six rectangular panels, each 115 cm 235 cm,
each with a blocked rectangular area (needed for connections) measuring 12.0 cm 22.5 cm. Find
the total collecting area in square metres.
Estimate: If the panels were 1.2 m by 2.4 m, or 2.9 m2 each, six of them would be 6 2.9, or
17.4 m2. Since we have estimated each dimension higher than it really is, and have also ignored
the blocked areas, we would expect this answer to be higher than the actual area.
Solution:
area of each panel (115)(235) 27 000 cm2
bocked area (12.0)(22.5) 270 cm2
Subtracting yields
collecting area per panel 27 000 270 26 700 cm2
There are six panels, so
total collecting area (6)26 700 160 000 cm2
16.0 m2
since there are 100 100 104 square centimetres in a square metre. This is lower than our
◆◆◆
estimate, as expected.