Download Triangles - BCIT Commons

The Mathematics 11
Competency Test
Triangles
Triangles are plane geometric figures with three straight line
sides and three vertices. Typically, the three vertices are
labelled with upper case characters from the beginning of
the alphabet, and the sides opposite each vertex are
labelled with the corresponding lower case letters, as shown
in the sketch to the right.
A
b
c
C
Often, the symbol ‘∆’ is used for the word ‘triangle.” Specific
B
a
triangles are identified by listing the characters labelling
their vertices. Thus ‘∆ABC’ is a shorthand notation to denote the triangle sketched above to the
right.
Types of Triangles
There are quite a number of terms to describe triangles of various
special types. You should be familiar with at least the following types:
600
equilateral triangles: all three sides have the same length. All three
vertices are the same size: 600. Notice how sometimes small crosslines are used to indicate which sides of the triangle have equal
lengths.
600
600
isosceles triangles: two of the three sides have the same length. The
two angles opposite the equal-length sides have the same measure.
scalene triangle: this term is used to refer to a general triangle, having no specific relationships
between the lengths of any sides. One thing we can say about the lengths of the sides of any
triangle in general is that no one side can be longer than the sum of the lengths of the other two
sides.
right triangle: has one vertex which forms a right
angle. The vertex with the right angle is
conventionally labelled C. The two shorter
perpendicular sides are called legs of the right
triangle. The longest side, opposite the right angle,
is called the hypotenuse.
A very important property of right triangles is
Pythagoras’s formula or theorem:
A
b
c
C
leg
a
B
c 2 = a2 + b 2
This formula will be exploited later when we introduce trigonometry.
David W. Sabo (2003)
Triangles
Page 1 of 5
Properties of Triangles
For every plane triangle, the vertex angles always add up to 1800:
A + B + C = 1800
Example: Two angles in a triangle are 63.60 and 42.10. Determine the third angle.
solution:
Let x be the third angle which we are to determine. Then, since all three angles must add up to
1800, we must have that
x + 63.60 + 42.10 = 1800.
So
x = 1800 – (63.60 + 42.10) = 74.30.
The third angle must be 74.30.
Example: Determine the measure of angles A and B in the
diagram to the right.
solution:
The three angles inside the triangle must add up to 1800.
The angle at the lower right is indicated to be a right angle,
so it must measure 900. Thus
A
B
380
380 + 900 + A = 1800
Therefore,
A = 1800 – (380 + 900) = 520
Angle B, together with the 380 angle form a straight angle which as a measure of 1800. Thus
B + 380 = 1800
So
B = 1800 – 380 = 1420.
For any right triangle, C = 900, so A + B = 900 – the two acute angles are complementary.
In any triangle, the longest side is always opposite the largest angle, and the shortest side is
always opposite the smallest angle.
Perimeter of a Triangle
The perimeter, P, or distance around any triangle, is simply the sum of the lengths of its sides:
P =a+b+c
David W. Sabo (2003)
Triangles
Page 2 of 5
Example: A triangular plot of land is sketched in the
figure to the right. Compute its perimeter and the cost of
putting a fence around it if the fence material costs
$15.95 per meter.
solution: The perimeter is just the sum of the lengths of
the three sides:
P = 42.6 m + 81.3 m + 102.7 m
= 226.6 m
To get the cost of the fence, just multiply the number of meters of fence required (the perimeter of
the triangle) by the cost in dollars per meter. Thus
$

Cost = 226.6 m ×  15.95  = $3614.27 .
m

Area of a Triangle
Imagine rotating a triangle so that one of its sides is
horizontal, as shown in the sketch to the right. We will
call this side the base of the triangle. Now, draw in a
altitude
vertical line from the top vertex of the triangle to this
(h)
base. Such a line is called an altitude of the triangle.
Obviously, since this altitude is vertical and this base
is horizontal, they are perpendicular. If we symbolize
base (b)
the length of the base of the triangle by the letter b,
and symbolize the length of the corresponding altitude
by the letter h (for “height”), then the area, A, of the triangle, the amount of surface it occupies, is
given by
A=
1
bh
2
(one-half base times height)
You will get exactly the same answer if you use any of
the three sides of the triangle as the base. In some
cases, the altitude may not be inside the triangle, as
was the situation in the sketch above. Instead, it may
have to be a vertical line from the top vertex drawn to
intersect an extension of the base, and shown to the
right. The base is still just the actual horizontal side of
the triangle and not its extension.
David W. Sabo (2003)
Triangles
base (b)
Page 3 of 5
Example: Calculate the perimeter and the area of the triangle
shown in the sketch to the right.
solution: This triangle is a right triangle with the length of the two
legs given. To calculate the perimeter, we need the lengths of all
three sides. Fortunately, since this is a right triangle, Pythagoras’s
theorem applies, giving us a way to calculate the length of the third
side (in this case, the hypotenuse).
So, letting c stand for the length of the longest side of this triangle,
we have by Pythagoras that
c2 = (8 m)2 + (15 m)2
= 64 m2 + 225 m2 = 289 m2
c
Therefore,
c = 289 m 2 = 17 m
So,
perimeter = p = sum of the lengths of the three sides
= 8 m + 15 m + 17 m = 40 m.
To calculate the area, imagine rotating the triangle so that the
side of length 15 m is on the bottom, horizontal. Then the side
of length 8 m will be vertical (since it is perpendicular to the side
of length 15 m). In this orientation, the side of length 15 m forms
a base of the triangle, (so b = 15 m) and the side of length 8 m
forms the corresponding altitude of the triangle (h = 8 m). Thus,
we get
Area = A =
8m
15 m
1
bh
2
=
1
(15 m )( 8 m ) = 60 m 2
2
Example: With reference to the figure to the right:
C B
(i) compute the perimeter and area of the triangle drawn with
heavy solid lines, and,
(ii) determine the measures of the angles labelled A, B,
and C.
A
solution:
To calculate the perimeter and areas, we need to deal
with lengths. So, we’ll resketch this triangle, showing all
lengths and rotating it so that the lower right hand side
is horizontal.
David W. Sabo (2003)
Triangles
Page 4 of 5
To compute the perimeter, we need the lengths of all three
sides. Two are already given. To get the length of the
third side, labelled ‘b’ in this rotated sketch, we note that
b + 12.3 m = 46.8 m
Therefore
12.3 m
b
46.8 m
b = 46.8 m – 12.3 m = 34.5 m.
So, now,
perimeter = p = 34.5 m + 22.4 m + 50.4 m = 107.3 m.
To calculate the area, we note that the vertical length, 18.7 m, forms an altitude (h = 18.7 m)
corresponding to the base, b = 34.5 m, formed by the horizontal side in the rotated sketch. Thus
Area = A =
1
1
bh = ( 34.5 m )(18.7 m ) = 322.575 m 2
2
2
To determine the measures of the angles A, B, and C, we make use of the fact that the sum of
the angles inside a triangle is always 1800, and angles combining to form a straight angle also
sum to 1800. Thus
and
so
B + 900 + 56.70 = 1800
⇒
B = 33.30
A + 56.70 = 1800
⇒
A = 123.30
C + A + 21.80 = 1800
C = 1800 – 21.80 – A
= 1800 – 21.80 – 123.30
= 34.90
David W. Sabo (2003)
Triangles
Page 5 of 5