Download National Protestant College Mathematics

National Protestant College
Grade 7A&B
Revision
Mathematics
Sheet Three
Triangles
Part I:
Triangles
1) A triangle is a polygon having three sides.
A
B
C
(Fig: 1)
In any triangle we have three sides, three angles and three vertices.
 ABC
is opposite to side [AC] or [AC] is opposite to side
 ABC
 BAC is opposite to side [BC] or [BC] is opposite to side  BAC
 ACB is opposite to side [AB] or [AB] is opposite to side  ACB
2) The sum of angles in any triangle is equal to 180  .
If  ABC= 70  ,  BAC= 50  then  ACB=




180    ABC   BAC   180   70   50   180   120   60 
Part II: Classification of triangles according to angles
1) Acute triangle is a triangle having three acute angles
A
B
C
(Fig: 2)
2) A right triangle is a triangle having one right angle. The side opposite to the right
angle is called hypotenuse and the other two sides are called legs.
P
Hypotenuse
leg
M
N
(Fig: 3)
leg
Triangle MNP is right angled at M
1
3) Obtuse triangle is a triangle having one obtuse angles
R
S
T (Fig: 4)
Triangle RST is an obtuse triangle since  RST=120  .
Part IV: Classification of triangles according to sides
1) Scalene triangle is a triangle having three none equal sides.
A
B
C
AB = 3.8 cm; BC = 5cm; CA = 2.3 cm
(Fig: 5)
2) An isosceles triangle is a triangle having two equal sides. The third side is called
the base and the angle at the base are called base angles. The base angles in any
isosceles triangle are equal.
N
M
L
(Fig: 6)
Triangle MNL is an isosceles triangle with vertex N means NM = NL
3) An equilateral triangle is a triangle having three equal sides and three equal
angles each measuring 60  .
A
B
C
(Fig: 7)
2
Part V: Special triangles
1) A right triangle having an angle of 60  and an angle of 30  is called halfequilateral triangle.
A
30 
B
C
60 
(Fig: 8)
2) A right triangle having two equal sides is called right isosceles. The base angles
measure each 45  .
B
45 
A
Part IV:
C
45 
(Fig: 9)
Important Lines in Triangles
Definitions:
1) The median in any triangle is the segment joining the vertex to the midpoint of the
opposite side. The three medians in any triangle are concurrent at a point called
center of gravity or centroid.
A
Center of gravity
P
N
(Fig: 10)
B
[AM] is the median relative to [BC]
C
M
[BN] is the median relative to [AC]
[CP] is the median relative to [BA]
2) The altitude or height in any triangle is the perpendicular dropped from the
vertex to the line holding the opposite side. The three altitudes in any triangle are
concurrent at a point called orthocenter.
3
Acute triangle
A
R
K
Orthocenter
B
C
(Fig: 11)
I
[AI] is the altitude relative to [BC]
[BR] is the altitude relative to [AC]
[CK] is the altitude relative to [BA]
In any acute triangle the orthocenter is inside the triangle
Right triangle
A
H
Orthocenter
B
C
(Fig: 12)
[AB] is the altitude relative to [BC]
[BC] is the altitude relative to [AB]
[CH] is the altitude relative to [AC]
In any right triangle the orthocenter is the vertex of the right angle
4
Obtuse triangle
A
K
H
C
(Fig: 13)
B
S
Orthocenter
[AH] is the altitude relative to the line holding [BC]
[BK] is the altitude relative to [AC]
[CS] is the altitude relative to the line holding [BA]
In any obtuse triangle the orthocenter is outside the triangle
3) The bisector in any triangle is the bisector of any angle of this triangle. The three
bisectors of the three angles in any triangle are concurrent at a point called center
of the inscribed circle or the in-center.
In-center
(Fig: 14)
4) The three perpendicular bisectors of the three sides in any triangle are concurrent
at a point called center of the circumscribed circle or the circum-center.
A
Circum-center
.
O
B
C
(Fig: 15)
5
In any right triangle the circum-center is the midpoint of the
hypotenuse
Part V:
Construction of Triangles
1) Construction of triangles knowing three sides
Example: Construct a triangle ABC such that AB=5cm, AC=4cm and BC=6cm.
Step 1: Draw BC=6cm.
Step 2: With center B and with a radius equal to 5cm draw an arc in the upper semiplane
Step 3: With center C and with a radius equal to 4cm draw an arc in the upper semiplane cutting the first arc in A.
Step 4: Join A to B and A to C. Write the coded properties.
6
2) Construction of triangles knowing two sides and the angle
included between these two sides.
Example: Construct a triangle MNP such that MN=5cm, MP= 4cm and  PMN = 70  .
Step 1: Draw  xMy = 70 
Step 2: With center M and with a radius equal to 5cm draw an arc cutting [Mx) in N
Step 3: With center M and with a radius equal to 4cm draw an arc cutting [My) in P.
Step 4: Join P to N. Write the coded properties
7
2) Construction of triangles knowing two angles and the side
holding these two angles.
Example: Construct a triangle RST such that RS = 6cm,  TRS = 70  and  RST =
40  .
Step 1: Draw RS = 6cm
Step 2: Draw [Rx) in the upper semi-plane such that  xRS = 70 
Step 3: Draw [Sy) in the upper semi-plane such that  RSy = 40 
Step 4: Point T is the point of intersection of [Rx) and [Sy).Write the coded properties
8
2) Construction of triangles knowing two sides and the angle
included between these two sides.
Example: Construct a triangle MNP such that MN=6cm, MP= 5cm and  PMN = 80  .
Step 1: Draw  xMy = 80 
Step 2: With center M and with a radius equal to 6cm draw an arc cutting [Mx) in N
Step 3: With center M and with a radius equal to 5cm draw an arc cutting [My) in P.
Step 4: Join P to N. Write the coded properties
9
Important Theorems in Triangles
1) The sum of angles in a triangles is equal to 180 
2) Property of sides in any triangle: In any triangle the length of any side is
greater than the sum and less than the difference of the other two sides.
b
c
a
Note:
If a>b>c then
 a-b<c<a+b
 a-c<b<a+c
 b-c<a<b+c
Examples:
a) If ABC is a triangle such that AB = 5 cm , AC = 4cm and BC = 3 cm then
AB-AC<BC<AB+AC  5-4< BC < 5+4
and AB-BC<AC<AB+BC  5-3< AC < 5+3
AC-BC<AB<AC+BC  4-3< BC < 4+3
b) If ABC is a triangle such that AB = 10 cm , AC = 8cm and BC = 6 cm then
AB-AC<BC<AB+AC  10-8< BC < 10+8
and AB-BC<AC<AB+BC  10-6< AC < 10+6
AC-BC<AB<AC+BC  8-6< AB < 8+6
c) Can you draw a triangle ABC such that AB = 6cm , AC = 4 cm and BC = 12
cm? Why?
3)
If two angles in one triangle are equal to two angles in another
triangle then the third pair of angles are equal
M
Given:  A =  M
A
B = N
Required to prove:  C =  P
B
C
N
P
Proof:
 A+  B+  C = 180  (sum of angles in any triangle)
 M+  N+  P = 180  (sum of angles in any triangle
  A+  B+  C =  M+  N+  P (transitive property)
But  A =  M and  B =  N (given)
  M+  N+  C =  M+  N+  P (by substitution)
  C =  P (subtracting  M and  N from both sides)
10
4) The acute angle in any right triangle are complementary
Given: Triangle ABC is a right triangle at A
Required to prove:  C +  B = 90 
Proof:
 A+  C +  B = 180  (sum of angles in any triangle )
 A = 90  ( since triangle ABC is right at A)
90  +  C +  B = 180  (by substitution)
 90  +  C +  B -90  = 180  -90  (subtracting 90  from both sides)
  C +  B = 90 
5) Note 1:
Exterior angle
Interior angle
6) The exterior angle in any triangle is equal to the sum of the opposite
interior angles
Given: Triangle ABC
 A 2 is the exterior angle at A
Required to prove:  A 2 =  B +  C
Proof:
B

 A +  B +  C =180 (sum of angles in any triangle)
 A 1 +  B+  C=  A 1 +  A 2 (transitive property)
 A 1 =  A 1 (reflexive property)
 A

1
2
1 A
C
+  B+  C-  A 1 =  A 1 +  A 2 -  A 1 (subtraction property of equality)
 A 2 =  B +  C (subtracting  A 1 from both sides)
7) Note 2: The exterior bisector is the bisector of the exterior angle.
C
x
y
Exterior bisector
Interior bisector
A
B
11
8) The angle formed by the interior and the exterior bisector form a right
angle.
y
Given: Triangle ABC
2
[Ax) is the interior bisector of  A
A
[Ay) is the exterior bisector of  A
1
Required to prove:  xAy =90 
B
C
x
9) If two angles of a triangle are congruent, then the sides opposite to these
two angles are also congruent.
10)
Triangular Inequality
Case 1:
If point B belongs to line (AC) such that B is between A and C then AC=AB+BC
A
B
C
Case 2:
If point B does not belong to line (AC) then AC  AB+BC
A
Case 3:
If point B belongs to line (AC) then AC  AB+BC
A
For all points A, B, C
AC 
C
(Fig: 22)
AB+BC
B
(Fig: 21)
C
B
12