plane geometry
DESCRIPTION
Plane Geometry for High SchoolTRANSCRIPT
CHAPTER 1: BASIC CONCEPTS IN GEOMETRY
1.1 Points, Lines & Planes
The most fundamental geometric form is a point. It is represented as a dot with a
capital alphabet which is its name (Figure 1.1) A line is a set of points and it
extends in opposite directions up to infinity. It is represented by two points on the
line and a double headed arrow or a single alphabet in the lower case (Figure 1.1)
A plane is a two dimensional (flat) surface that extends in all directions up to
infinity.
A plane has obviously no size and definitely no shape. However it is represented as
a quadrangle and a single capital letter (Figure 1.1)
Figure 1.1 shows points A, D & Q, line AB, line l and plane P.
Some axioms regarding points, lines and planes are given below.
1. An infinite number of lines can be drawn through any given point.
2. One and only one line can be drawn through two distinct points.
3. When two lines intersect they do so at only one point.
Collinear and Coplanar
Three or more points are said to be collinear if a single line contains all of them.
Otherwise they are said to be non collinear. (Figure 1.2)
Figure 1.2
Figure 1.2 shows two lines l and m. Line l is such that it passes through A, B and
C. Hence points A B and C are collinear. In the case of points P, Q and R there can
be no single line containing all three of them hence they are called non-linear.
Similarly points and lines which lie in the same plane are called coplanar otherwise
they are called non-coplanar.
Axiom: A plane containing a line and a point outside it or by using the definition
of a line, a plane can be said to contain three non-collinear points. Conversely,
through any three non collinear points there can be one and only one plane (figure
1.3).
Axiom: If two lines intersect, exactly one plane passes through both of them
(figure1.4).
Axiom: If two planes intersect their intersection is exactly one line (figure 1.5).
Figure 1.3
Lines A, B and C are contained in the same plane P or A, B and C are three non-
collinear points through which one plain P can pass.
Figure 1.4
Plane Q contains intersecting lines l and m.
Figure 1.5
Planes P and Q intersect and their intersection is line l.
Axiom: If a line does not lie in a plane but intersects it, their intersection is a point
(figure 1.7 ).
Figure 1.6
Point A is the intersection point of line l and plane P.
Example 1
Take any three non-collinear points A,B and C on a paper. How many different
lines can be drawn through different pairs of points ? Name the lines.
Solution:
Three lines can be drawn namely AB, BC & AC.
Example 2
Figure 1.7
From figure 1.7 answer the following:
a. Name lines parallel to line AB
b. Are line AO and point R coplanar ? Why ?
c. Are points A, S, B and R coplanar ? Why ?
d. Name three planes passing through at A.
Solution:
a. Line CD, line SR and line PQ.
b. Yes. Any line and a point outside it are coplanar.
c. Yes. Two parallel lines are always coplanar.
d. ABCD, ADSP and ADCB.
1.2 Line Segment
A line segment is a part of a line. It has a fixed length and consequently two end
points. They are used to name the line segment (figure 1.8).
Figure 1.8
Seg. PQ is a segment of line AB. A line contains infinite segments and if two
segments on a line have a common end point, they can be added ( figure 1.9).
Figure 1.9
Seg. PQ and Seg.QR are two segments on line l and they have a common end point
Q. Therefore Seg.PQ + Seg.QR = Seg.PR.
On segment Seg.PR there exists a point M. Therefore seg. PM + seg.MR = seg.PR
or seg. PM = seg. MR , it implies that seg.PR = 2 seg. PM = 2 seg. MR. In other
words it means that M is equidistant from P & R. Therefore, M is the midpoint of
seg.PR. Every segment has one and only one midpoint.
Exercise
Are the following statements true or false ?
1. Any number of lines can pass through a single given point.
2. If two points lie in a plane the line joining them also lies in the same plane.
3. Any number of lines can pass through two given points.
4. Two lines can intersect in more than one point.
5. Two planes intersect to give two lines.
6. If two lines intersect only one plane contains both the lines.
7. A line segment has two end points and hence a fixed length.
8. The distance of the midpoint of a segment from one end may or may not be
equal to its distance from the other.
Answers
1. True
2. True
3. False. One and only one line passes through two given points.
4. False. Two line can intersect in only one point.
5. False. Two planes intersect to give one line.
6. True
7. True
8. False. The midpoint is equidistant from both ends of a segment.
1.3 Rays and Angles
A ray has one end point and extends in the other direction upto infinity. It is
represented by naming the end point and any other point on the ray with the
symbol(figure 1.10 ).
Figure 1.10
J is the end point of a ray and K is a point on it. This ray is represented as . A
ray can extend in any one direction only.
Two rays going in different directions, but having a common end point, form an
angle. The common end point is called the vertex of the angle and the rays are
called its sides or arms. An angle is represented by the symbol and named, using
either both the rays or just the vertex (figure 1.11).
Figure 1.11
Figure 1.11 represents XYZ or Y.
Interior and Exterior of an angle: The interior of PQR is the shaded region in
figure 1.12. S is a point in the interior of Q because it lies on the R- side of ray
PQ and the P - side of ray QR. The set of all such points is called the interior
of PQR.
Figure 1.12
In figure 1.13 the shaded region shows the exterior of XYZ. The exterior of an
angle is defined as the set of points in the plane of a given angle which are neither
on the sides of the angle nor in its interior.
Figure 1.13
Measure of an angle: Every angle has a measure. It is measured in degrees from
00to 1800 and is represented as m . A line is also an angle because it satisfies the
definition of having two rays going in different ( in this case opposite ) directions
with a common end point ( figure 1.14 ).
Figure 1.14
is also AOB and m AOB = 1800
Figure 1.15
All rays starting from O and going above line AB form angles with such that
their measure is between 00 and 1800. eg. COB , DOB and EOB.
Angle addition property: Two angles with a common side and a common vertex
are adjacent if their interiors are disjoint. The measures of two adjacent angles can
be added to find the measure of the resultant angle. This is called the angle
addition property. With reference to figure 1.15 m COB + m DOC =
m DOB.
Angle Bisectors: The ray which passes through the vertex on an angle and divides
the angle into two angles of equal measure is called the bisector of that angle. Of
the two angles (figure 1.16 ) AOB and COB are equal in measure then is
called the bisector of COA. Just as every line has only one midpoint every angle
has only one bisector.
Figure 1.16
Types of angles: Depending on their measure, angles can be classified as acute
angle, right angle or obtuse angle.
Right angle: The right angle (figure 1.17 ) has a measure of 900. It is represented
by the symbol . Since its measure is fixed, it follows that all right angles are
equal. AOB (figure 1.17) is a right angle.
Figure 1.17
Acute angle: Any angle whose measure is between 0 and 900 is called an acute
angle (figure 1.18).
Figure 1.18
0 < a < 90 a is an acute angle.
Obtuse angle: An angle with a measure between 00 and 1800 is called an obtuse
angle (figure 1.19).
Figure 1.19
900 < b < 1800 b is an obtuse angle.
1.4 Some special angles
Complementary angles: If the sum of two angles equals 900 the two angles are
called complementary. Complementary angles thus add up to a right angle.
Complementary angles are of two types. If they have one side in common they are
called adjacent complementary angles (figure 1.20 a). If no side is common then
they are called non-adjacent complementary angles (figure 1.20 b).
Figure 1.20 a Figure 1.20 b
Since the measures of complementary angles always sum up to 900 if the measure
of one angle is known that of the complement can be found easily. In figure 1.20
b a and b are complementary. Also it is known that m a = 300.
m a + m b = 900
300 + m b = 900
m b = 900 300
m b = 600
Theorem: If two angles are complementary to a third angle, then they are equal to
each other.
Proof: a and b are both complementary to c.
m a + m c = 900 and also m b + m c = 900
m a + m c = m b + m c OR m a = m b
Theorem: If two angles are complementary to equal angles they are equal to each
other . If a and b are complementary to c and d respectively where m c = m d.
Proof: m a + m c = m b + m d = 900
m a + m c = m b + m d Since m c = m d
m a + m c = m b + m c or m a = m b.
Supplementary angles: If the measures of two angles sum up to 1800 they are
called supplementary angles. Supplementary angles are of two types:
a) Non adjacent supplementary angles and
b) Adjacent supplementary angles.
Non adjacent supplementary angles are distinct and have no arm in common
(figure 1.21).
Figure 1.21
Ð A and B are supplementary and non adjacent.
Adjacent supplementary angles are called angles in a linear pair and have one arm
in common (figure 1.22).
Figure 1.22
Vertical angles: When two lines AB and CD intersect at O, four angles are formed
with vertex O. Consider AOC and BOD. It is observed that and are
opposite rays and so is and . In such a case AOC and BOD are called
vertical angles (figure 1.23 ).
Figure 1.23
Theorem: Vertical angles are always equal in measure.
Proof: To prove m AOC = m BOD
m AOC + m COB = 1800 ( supplementary angles )
m BOD + m COB = 1800 ( supplementary angles )
i.e. m AOC + m COB = m BOD + m COB
or m AOC = m BOD.
Example 1
Measures of some angles are given below. Find the measures of their supplements.
a) 750 b) 1250 c) x0 d) (180 - x) 0 e) (90 + x) 0
Solution:
a) 1050 b) 550 c) (180 - x) 0 d) x0 e) (90 - x) 0
Example 2
Measures of some angles are given below. Find the measures of their
complements.
a) 350 b) 450 c) (90 - r) 0 d) x0
Solution:
a) 550 b) 450 c) r0 d) (90 - x) 0
Example: 3
The measure of one angle is twice that of its complement. Find its measure.
Solution: 600
Example 4
The measure of an angle is four times the measure of its supplementary angle. Find
its measure.
Solution: 1440
1.5 Angles made by a transversal
Definition of a transversal: A line which intersects two or more given coplanar
lines in distinct points is called a transversal of the given lines. In figure 1.24 the
line l is the transversal of lines a and b.
Figure 1.24
l intersects a and b at P and Q respectively. The three lines determine eight angles
four with vertex P and the remaining four with vertex Q.
Corresponding angles: Angles that appear in the same relative position in each
group are called corresponding angles, i.e. 1 and 5 are called corresponding
angles. Similarly 2 & 6, 4 & 8 and 3 & 7 are pairs of corresponding
angles.
Interior and Exterior angles: Those angles which lie between lines a and b are
called interior angles, i.e. 3, , 5and Exterior angles lie on opposite
sides of lines a and b, i.e. and Ð 8.
Alternate and Consecutive Interior angles: Interior angles on opposite sides of
the transversal are called alternate interior angles and , are alternate
interior angles and so also 3, and .
Interior angles on same side of the transversal are called consecutive interior
angles. , and are consecutive interior angles and so also 3 and Ð 6.
Alternate and Consecutive Exterior angles: Alternate exterior angles are on
opposite sides of the transversal and do not lie between lines a and b,
i.e. andand also 2 and .
Exterior angles on the same side of the transversal are called consecutive exterior
angles, i.e. 8as also 2 and .
Example 1
In figure 1.25 n is the transversal of lines l and m. Write down the pairs of:
a) corresponding angles,
b) alternate interior angles,
c) alternate exterior angles ,
d) consecutive interior angles & ,
e) consecutive exterior angles.
Figure 1.25
Answers
a) a & e , b & f , c & g , d & h .
b) d & f c & e
c) a & g , b & h
d) f & c , d & e
e) a & h , b & g .
1.6 Transversal across two parallel lines
Corresponding Angles: If a transversal cuts two parallel lines the corresponding
angles are equal (figure 1.26)
Figure 1.26
l & m are two parallel lines cut by a transversal n to form angles 1 to 8.
Axiom: Corresponding angles are equal in measure if a transversal cuts parallel
lines.
m 1 = m 5
m 2 = m 6
m 3 = m 7
m 4 = m 8 .
Alternate interior angles: If a transversal cuts two parallel lines the alternate
interior angles are equal in measure. In figure 1.26 m 4 = m 6 and m 3 =
m 5. This can be proved as follows:
m 3 = m 4 are supplementary and so also
m 6 = m 7 are supplementary.
Since the sums of their measures are 1800 in both cases.
m 3 + m 4 = m 6 + m 7.
However m 3 = m 7 as they are corresponding angles formed by a transversal
across parallel lines.
Therefore, m 3 + m 4 = m 6 + m 3 i.e. m 4 = m 6.
Similarly it can be shown that m 3 = m 5 .
Alternate exterior angles: If a transversal cuts two parallel lines the alternate
exterior angles are equal in measure. In figure 1.26 m 1= m 7 and m 2 =
m 8. This can be easily shown as follows:
m 1 = m 3 (vertical angles)
m 3 = m 7 (corresponding angles on parallel lines)
m 1 = m 7
Consecutive Interior and Consecutive Exterior Angles: If a transversal cuts two
parallel lines the consecutive interior angles and consecutive exterior angles are
supplementary. In figure 1.26 m 4 and m 5 are supplementary and also m 3
and m 6 are supplementary. This is proven as:
m 1 and m 4 are supplementary
but m 1 = m 5 (corresponding angles on parallel lines )
m 4 and m 5 are supplementary.
Since consecutive exterior angles are supplementary. m 1 and m 8 so also
m 2 and m 7 should be supplementary.
Proof: m and m 4 are supplementary but
m = m 8 (corresponding angles on parallel lines )
m 1 and m 8 are supplementary.
1.7 Conditions for parallelism
Parallel lines are defined as those lines which are coplanar and do not intersect
(figure 1.27). The figure merely shows a part of the lines as the lines actually
extend upto infinity. Hence although they do not intersect in the region that is
observed it is possible that they will intersect when sufficiently extended.
In practice however the lines cannot be extended beyond the paper. Therefore it is
not convenient to test parallelism by looking for a point of intersection.
Figure 1.27
If however a transversal is drawn across lines l and m figure 1.28 all the eight
angles formed can be measured.
Figure 1.28
By using properties of these angles that are given in section 1.6 line l and m can be
shown to be parallel.
Lines l and m are parallel if:
1) corresponding angles formed by line n are equal in measure,
2) alternate interior angles are equal in measures,
3) the measures of alternate exterior angles are equal,
4) consecutive interior angles are supplementary,
5) consecutive exterior angles are supplementary.
These are called conditions for parallelism.
Example 1
Lines l and m are parallel and line n is the transversal find the measures of angles
a, b, c, d, e, f and g.
Figure 1.29
Solution :
m a = 1050 m b = 750
m c = 1050 m d = 1050
m e = 750 m f = 1050
m g = 750
Example 2
Figure 1.30
Prove that is parallel to .
Solution:
PR is a transversal to lines containing m QPR = m PRS = 500
Since they are alternate interior angles is parallel to .
Therefore SRT and RTU are supplementary angles formed by the transversal
TR on line containing . Therefore is parallel to . It is already
shown that is parallel to . Therefore is parallel to .
**********
CHAPTER 2: TRIANGLES
2.1 Introduction
As the name triangle suggests, this geometric shape is made of three angles. It has
three sides and is represented by the symbol and is named by its vertices as
shown in figure 2.1.
Figure 2.1
ABC has three sides AB, BC and CA.
It has three angles ABC, BCA and CAB.
2.2 Sum of the angles of a triangle
It can be proven easily that the sum of the three angles of a is 1800
Figure 2.2
ABC in figure 2.2 is a triangle with line l parallel to seg. BC and passing through
A. seg. AB is a transversal on two parallel lines seg. PQ and seg.BC. Hence
m PAB and m ABC are equal as they are alternate interior angles. Similarly
m QAC = m ACB.
Now PAQ = m PAB + m BAC + m CAQ
i.e. 1800 = m ABC + m BAC + m ACB
m PAQ = 1800 because it is a straight line. Thus the sum, of the measures of the
three angles, of any triangle, is 1800.
2.3 Types of triangles
Triangles are classified into various types, using two different parameters - the
lengths of their sides and the measure of their angles.
Length of the Side
Based on the lengths of their sides, triangles are classified into three categories.
a. Equilateral triangle: If the lengths of all three sides of the triangle are
equal, then it is called an equilateral triangle. Figure 2.3 shows an equilateral
triangle.
Figure 2.3
b. Isosceles triangle: If only two sides of a triangle are equal in length, it is
called as an isosceles triangle. Figure 2.4 shows an isosceles triangle.
Figure 2.4
c. Scalene triangle: If all the sides of a triangle have different lengths it is
called a scalene triangle. Figure 2.5 shows a scalene triangle.
Figure 2.5
Angles
a. Acute triangle: A triangle in which all the angles are acute, ( i.e. < 900 ) is
called as an acute triangle. Figure 2.6 shows an acute triangle.
Figure 2.6
A special case of an acute triangle is when all the three acute angles are
equal. This is called an equiangular triangle. Figure 2.7 shows an
equiangular triangle.
Figure 2.7
Since the sum of all the angles of a triangle is 1800, it can be said that each
angle of an equiangular triangle is 600.
b. Obtuse triangle: A triangle in which one of the angles is obtuse is called as
an obtuse triangle. Figure 2.8 shows an obtuse triangle.
Figure 2.8
Since the sum of all the angles of a triangle is 1800 it can be said that the
other two angles of an obtuse triangle are acute.
c. Right Triangle: It is a triangle in which one of the angles is a right angle.
Figure 2.9 shows a right triangle.
Figure 2.9
Since KJL is 900 it can be said that JKJL and JLK are
complementary. In a right triangle the side opposite to the right angle is
called the hypotenuse.
2.4 Altitude, Median and Angle Bisector
Altitude
An altitude is a perpendicular dropped from one vertex to the side (or its extension)
opposite to the vertex. It measures the distance between the vertex and the line
which is the opposite side. Since every triangle has three vertices it has three
altitudes.
a. Altitudes of an acute triangle:
Figure 2.10
For an acute triangle figure 2.10 all the altitudes are present in the triangle.
b. Altitudes for a right triangle:
Figure 2.11
For a right triangle two of the altitudes lie on the sides of the triangle, seg.
AB is an altitude from A on to seg. BC and seg. CB is an altitude from C on
to seg.AB. Both of them are on the sides of the triangle. The third altitude is
seg. BD i.e.from B on to AC. The intersection point of seg. AB, seg. BC and
seg. BD is B. Thus for a right triangle the three altitudes intersect at the
vertex of the right angle.
c. Altitudes for an obtuse triangle:
Figure 2.12
ABC is an obtuse triangle. Altitude from A meets line containing seg.BC
at D. Therefore seg. AD is the altitude. Similarly seg.CE is altitude on to AB
and BF is the altitude on to seg. AC. Of the three altitudes, only one is
present inside the triangle. The other two are on the extensions of line
containing the opposite side. These three altitudes meet at point P which is
outside the triangle.
Median
A line segment from the vertex of a triangle to the midpoint of the side opposite to
it is called a median. Thus every triangle has three medians. Figure 2.13 shows
medians for acute right and obtuse triangles.
Figure 2.13
All three medians always meet inside the triangle irrespective of the type of
triangle.
Angle Bisector
A line segment from the vertex to the opposite side such that it bisects the angle at
the vertex is called as angle bisector. Thus every triangle has three angle bisectors.
Figure 2.14 shows angle bisectors for acute right and obtuse triangles.
Figure 2.14
2.5 Congruence of triangles
Two triangles are said to be congruent, if all the corresponding parts are equal. The
symbol used for denoting congruence is and PQR STU implies that
and also
i.e. corresponding angles and corresponding sides are equal.
In order to prove that two triangles are congruent, it is not always necessary to
show that all the six corresponding parts are equal. If certain basic requirements are
met the triangles are said to be congruent. These basic criteria’s are embodied in
the five postulates given below. These postulates are axiomatic and are useful in
proving the congruence of triangles.
S S S Postulate
If all the sides of one triangle are congruent to the corresponding sides of another
triangle then the triangles are congruent (figure 2.15).
Figure 2.15
seg. AB = seg. PQ , seg. BC = seg. QR and
seg. CA = seg. RP
ABC PQR by S S S.
S A S Postulate
If the two sides and the angle included in one triangle are congruent to the
corresponding two sides and the angle included in another triangle then the two
triangles are congruent (figure 2.16).
Figure 2.16
seg. AB = seg. PQ , seg. BC = seg. QR and
m ABC = m PQR
ABC PQR by S A S postulate.
A S A Postulate
If two angles of one triangle and the side they include are congruent to the
corresponding angles and side of another triangle the two triangles are congruent
(figure 2.17).
Figure 2.17
m B + m R m L = m P and seg. BC = seg. RP
ABC QRP by A S A postulate.
A A S Postulate
If two angles of a triangle and a side not included by them are congruent to the
corresponding angles and side of another triangle the two triangles are congruent
(figure 2.18)
Figure 2.18
m A = m P m B = m Q and AC = PR
ABC PQR by A A S.
H S Postulate
This postulate is applicable only to right triangles. If the hypotenuse and any one
side of a right triangle are congruent to the hypotenuse and the corresponding side
of another right triangle then the two triangles are congruent (figure 2.19).
Figure 2.19
Then hypotenuse AC = hypotenuse PR
Side AB = Side PQ
ABC PQR by HS postulate.
Example 1
Name the postulate to be used to prove the following pairs of triangles are
congruent.
a)
b)
c)
d)
e)
f)
g)
Figure 2.20
Solution:
a) S S S b) S AS
c) A S A d) A A S
e) H L f) None as there is no AAA postulate.
g) None, there is no ASS postulate.
Example 2
In figure 2.21 seg. MN seg. MO , seg. PN seg. PO, prove that MNP
MOP.
Figure 2.21
Solution:
Draw segment MP
In triangle MNP and MOP
seg. MN seg.MO seg. PN seg. PO seg. MP seg. MP
MNP MOP ...(S S S Postulate)
MNP MOP ....(as they are corresponding angles of congruent triangles .)
Example 3
In figure 2.22 H and G are two points on congruent sides DE & DF of DEF such
that seg. DH seg. DG. Prove that seg. HF seg. GE.
Figure 2.22
Solution:
In triangles DHF and DGE seg. DF seg. DE
HDF GDE (same angle ) seg. DH seg. DG
DHF DGE S A S postulate seg. HF seg.
GE as they are corresponding sides of congruent triangles.
2.6 Sides opposite congruent angles
Theorem: If two sides of a triangle are equal, then the angles opposite them are
also equal. This can be proven as follows:
Consider a ABC where AB = AC (figure 2.23 ).
Figure 2.23
Proof: To prove m B = m C drop a median from A to BC at point P. Since AP
is the median, BP = CP.
In ABP and ACP
seg. AB seg. AC( given )
seg. BP seg. CP( P is midpoint )
seg. AP seg. AP( same line )
Therefore the two triangles are congruent by SSS postulate.
ABP ACP
m B = m C as they are corresponding angles of congruent triangles.
The converse of this theorem is also true and can be proven quite easily.
Consider ABC where m B = m C (figure 2.24 )
Figure 2.24
To prove AB = AC drop an angle bisector AP on to BC.
Since AP is a bisector m BAP = m CAP
m ABP = m ACP ( given )
seg. AP seg. AP (same side )
ABP ACP by AAS postulate.
Therefore the corresponding sides are equal.
seg. AB = seg. AC
Conclusion: If the two angles of a triangle are equal, then the sides opposite to
them are also equal.
**********
CHAPTER 3: POLYGONS
3.1 Definition
Polygons are closed planar shapes with three or more sides. Some examples of
polygons are given in figure 3.1.
Figure 3.1
3.2 Terminology
Vertices: The corners of the polygons are called vertices.
Consecutive sides: Consecutive sides are those which have a vertex in common.
Diagonals: Diagonals are segments joining non-consecutive vertices.
Figure 3.2
In figure 3.2 A, B, C, D, E & F are vertices. AB has two consecutive sides BC and
AF. Similarly two consecutive sides exist for the rest of the sides.
Segments joining A to all vertices except B & F are diagonals. Similarly, diagonals
can be drawn from all the other vertices.
3.3 Sum of Interior angles of a Polygon
Figure 3.3
Figure 3.3 shows an octagon. Five diagonals can be drawn from A. This gives rise
to six triangles.
Since the sum of all internal angles of a triangle is 1800, the sum all the internal
angles of this polygon is 6 1800 = 10800.
This can be generalized as:
For any n sided polygon the sum of its internal angles is ( n - 2 ) 1800.
3.4 Sum of exterior angles of a Polygon
Figure 3.4
Figure 3.4 shows a pentagon. Its external angles are named from a to e. The aim is
to find the sum of these five angles.
It is known that the sum of internal angles of a Pentagon
= (5 - 2 ) 1800
= 3 1800
= 540
each interior angle of the pentagon measures
5400 5 = 1080
The interior and exterior angles form linear pairs and hence are supplementary.
Each exterior angle measures 1800 - 1080 = 720
Sum of five exterior angles = 5 72 = 3600
It can be proved that the sum of the exterior angles for any polygon is 360 0.
Sum of interior angles of an n sided polygon = ( n - 2 ) 1800.
Measure of each internal angle =
Each exterior angle =
Sum of n exterior angles =
=
=
Conclusion: The sum of interior angles of a polygon is dependent on the number
of sides but the sum of the exterior angles is always 3600.
Example 1
Find the sum of the internal angles of a six, eight and ten sided polygon.
Solution: 7200, 10800, 14400
Example 2
Find the sum of the external angles of a twelve sided polygon.
Solution: 3600
Example 3
If the sum of the internal angles of a polygon is 12600 Find the number of sides.
Solution: Nine sides.
3.5 Trapezoids
A trapezoid is a four sided polygon, such that, one pair of opposite sides is parallel
to each other. Figure 3.5 shows a trapezoid ABCD.
Figure 3.5
The parallel sides of a trapezoid are called bases and the non parallel sides are
called legs. The line joining the mid points of the legs in a trapezoid is parallel to
the bases and is called the median.
Figure 3.6
In figure 3.6 ABCD is a trapezoid. P is the midpoint of AD and Q is the midpoint
of BC. Segment PQ is parallel to both AB and DC and is called the median of
ABCD. The median is always half of the sum of the bases,
i.e.
The distance between the bases ( parallel sides ) of a trapezoid is measured by the
perpendicular line joining the two bases. It is called the attitude of the trapezoid. In
Figure 3.7 AX and BY are the attitudes of trapezoid ABCD.
Figure 3.7
Consider a trapezoid figure 3.8 where the legs are equal in length. This is called as
an isosceles trapezoid.
Figure 3.8
In figure 3.8 seg.LO = seg.MN. Therefore, LMNO is an isosceles trapezoid. In
such a trapezoid the base angles are equal. This can be proven by drawing two
altitudes from L & M on the seg.ON.
Figure 3.9
Figure 3.9 shows LP & MQ as two altitudes of MNO.
Consider LOP & MNQ. Both are right triangles such that their hypotenuse has
the same length ( LMNO is an isosceles trapezoid ).
Also seg. LP = seg MQ as the perpendicular distance between two parallel lines is
always the same.
By HS postulate LOP MNQ.
LOP MNQ as corresponding angles of congruent triangles are congruent.
Another interesting property of an isosceles trapezoid is that the diagonals are
equal in length.
Figure 3.10
Figure 3.10 shows an isosceles trapezoid where LN & MO are the diagonals.
It can be easily proved that seg. LN = seg. MO. In figure 3.10 consider LON
& MNO.
seg. LO = seg. MN by definition of isosceles trapezoid.
LON MNO base angles are equal. seg. ON + seg. NO same side.
LON MNO by SAS postulate
segLN seg. MO as corresponding sides of congruent triangles are congruent.
Midpoint Theorem
The segment joining the midpoints of two sides of a triangle is parallel to the third
side and half as long as the third side. Recall that the median of a trapezoid is
parallel to both the bases and is half the sum of their lengths.
Figure 3.11
In figure 3.11 ABC can be considered to be like a trapezoid where one base is
BC and the other is point A. PQ, which is a median, is therefore parallel to BC and
is half its length.
Example 1
Figure 3.12
In figure 3.12 LMNO is an isosceles trapezoid. m MLO = 108 and seg. MO =
4.9. Find (a) m LMN , (b) m MNO and (c) seg. LN .
Solution:
a) m LMN = 1800 as the base angles of an isosceles trapezoid are equal.
b) m MNO = 720 The consecutive interior angles formed by a transversal seg.
MN on two parallel lines seg. LM & seg. NO are supplementary. Therefore LMN
and MNO are supplementary.
c) LN = 4.9 as the diagonals of an isosceles trapezoid are equal in length.
Example 2
Figure 3.13
In figure 3.13 PQRS is an isosceles trapezoid. seg. PQ = 30 and seg. SR = 50. Find
the length of the median XY.
Solution:
seg. XY = half the sum of the lengths of the bases
= ( 30 + 50 )
= ( 80 )
= 40
3.6 Parallelograms
Unlike trapezoids, which are quadrilaterals with only one pair of opposite sides as
parallel, if both the pairs of opposite sides are parallel, the quadrilateral is called a
parallelogram. Figure 3.14 is a parallelogram.
Figure 3.14
Seg.AB is parallel to seg.DC i.e. Seg. AB seg.DC and seg.AD is parallel to
seg.BC i.e. seg.AD seg.BC. Therefore ABCD is a parallelogram. It is represented
as parallelogram ABCD. Since both sides are parallel, a parallelogram has two
pairs of bases and hence two attitudes.
Properties of Parallelograms
Theorem: The opposite sides of a parallelogram are congruent.
Figure 3.15 shows a parallelogram ABCD to prove that seg.AB seg.CD &
seg.AD seg.BC.
Figure 3.15
Join A to C. Consider the two triangles ACB and CAD.
CAB ACD ( alternate angles )
ACB CAD ( alternate angles )
and seg.AC seg.CA ( same side )
ACB CAD ( ASA )
seg.AB seg.CD and seg.CB seg.DA as corresponding sides of congruent
angles are congruent.
Thus it is proved that the opposite sides of a parallelogram are congruent. From the
same proof it can be said that the diagonal of a parallelogram divides it into two
congruent triangles.
Since ACB CAD
ABC CDA
By drawing a diagonal from D to B it can be shown that DAB BCD which
means that in a parallelogram the opposite angles are congruent.
Another important feature of a parallelogram is given in the theorem below:
Theorem: The diagonals of a parallelogram bisect each other.
Figure 3.16 shows a parallelogram PQRS, seg.PR and seg.QS are its two diagonals
that intersect in O.
Figure 3.16
To prove that seg.PR & seg.QS bisect each other at O.
In SOR and QOP, OSR OQP and ORS OPQ (alternate angles ).
SR PQ opposite sides of a parallelogram.
SOR QOP ( ASA )
seg. SO seg.OQ i.e. O is the midpoint of SQ and seg.PO seg.OR i.e. O is the
midpoint of PR. Hence it is proved that PR and SQ bisect each other at O.
Summary of the properties of a Parallelogram
1) Both the pairs of opposite sides of a parallelogram are parallel to each other.
2) The opposite sides of a parallelogram are congruent.
3) The opposite angles of a parallelogram are congruent.
4) The two triangles, formed by a diagonal of a parallelogram, are congruent.
5) The diagonals of a parallelogram bisect each other.
Conditions for a parallelogram
The converse of the above theorems is proved below. These theorems give the
conditions under which a quadrilateral is a parallelogram.
Theorem: A quadrilateral is a parallelogram, if its opposite sides are congruent.
Figure 3.17 shows a quadrilateral ABCD with its opposite sides congruent.
Figure 3.17
To prove that ABCD is a parallelogram, join A to C and consider ADC
& CBA
AD CB and DC BA ( given ) also AC CA ( same side )
ADC CBA ( SSS ).
ACB CAD and ACD CAB ( corresponding angles of congruent
triangles are congruent ).
ACB CAD AD BC because they are alternate angles formed by the
transversal CD that intersects BC and AD. Since they are congruent the two lines
intersected by the transversal are parallel.
Similarly it can be show that since ACD CAB AB DC. Since both the
opposite sides are parallel to each other ABCD is a parallelogram.
Theorem: A quadrilateral is a parallelogram if it’s diagonals bisect each other.
Figure 3.18 shows a quadrilateral PQRS such that its diagonals seg.PR and seg.QS
bisect each other on the point of intersection O.
Figure 3.18
To prove that PQRS is a parallelogram, consider POS and ROQ.
seg.PO seg.RO and seg.OS seg.OQ ( given )
Ð POS ROQ ( vertical angles )
\D POS ROQ
\Ð OPS ORQ corresponding angles of congruent triangles are congruent.
These are alternate angles formed on seg.PS and seg.QR by the transversal seg.PR
and since they are congruent PS QR. Similarly by showing that OSR OQP,
it can be shown that PQ SR. Since both the pairs of opposite sides are parallel
lines PQRS is a parallelogram.
Theorem: If one pair of opposite sides is parallel and congruent, the quadrilateral
is a parallelogram.
In figure 3.19 there is a quadrilateral LMNO where seg.LM seg.NO and
seg.LMseg.NO . To prove that LMNO is a parallelogram.
Figure 3.19
Since seg.LM seg.ON LMO NOM ( alternate angles )
In LMO and NOM
seg.LM seg.NO Ð LMO NOM
seg.MO seg.OM (Same side)
\ D LMO NOM
\ Ð LOM NMO (corresponding angles of congruent triangles are congruent).
But they are alternate angles formed by seg.MO on seg.MN and seg.LO and since
they are congruent seg.LO seg.MN. Therefore LMNO is a parallelograme.
3.7 Square, Rectangle and Rhombus
These are special cases of parallelogram. Hence they have all the properties of a
parallelogram and some additional properties.
Rectangle
A parallelogram in which each angle is 900 is called a rectangle. Hence a rectangle has all the
properties of a parallelogram.
1) The opposite sides are parallel and congruent.
2) Diagonals bisect each other.
Apart from these the rectangle has one salient property.
Theorem: The diagonals of a rectangle are congruent.
Figure 3.20 shows a rectangle.
Figure 3.20
To prove that seg.AC seg.BD consider ACD and BDC . Both are right triangles.
seg.AD seg.BC by definition
seg.DC seg.CD same side
ADC BCD - both are right angles.
ACD BDC ( SAS )
Therefore, AC BD corresponding sides of congruent triangles are congruent. Therefore, the
diagonals of a rectangle are congruent The converse of this theorem is used as a test for
rectangle.
Theorem: A parallelogram is a rectangle, if its diagonals are congruent.
Figure 3.21 shows a parallelogram LMNO whose diagonals are congruent.
Figure 3.21
To prove that LMNO is a rectangle, consider LNO and MON.
seg.LN seg.MO given seg.LO seg.MN opposite sides of a parallelogram and
seg.ON seg.NO same side
LNO MON ( SSS )
LON MNO ( corresponding angles of congruent triangles ).
Since they are interior angles of parallel lines they are supplementary.
LON and MNO are both right angles. MNO is a rectangle.
Rhombus
A rhombus is defined as a quadrilateral with all sides congruent. Figure 3.20 shows a
rhombus. ABCD where seg.AB seg.BC seg.CD seg.DA. A rhombus has all properties
of a parallelogram and some more.
Additional properties of a rhombus
Theorem: The diagonals of a rhombus are perpendicular to each other.
Figure 3.20 shows a rhombus ABCD.
Figure 3.22
To prove that seg.AC is perpendicular to seg.BD, consider BOA and BOC.
seg.BA seg.BC definition of rhombus .......................... (1)
ABD CDB alternate angles. DBC CDB as seg.CD and seg.CB are
congruent. Therefore CBD is isosceles .
ABD DBC which is the same as ABO CBO ................. (2)
seg.BO seg.BO same side .......................... (3)
BOA BOC
BOA BOC.
Corresponding angles of congruent triangles are congruent.
But BOA and BOC form a linear pair. i.e. they are supplementary.
Supplementary angles are congruent if and only if they are right angles. Therefore,
AC is perpendicular to BD. The converse of this theorem is used as a test for
rhombus.
Theorem: If the diagonals of a parallelogram are perpendicular, the parallelogram
is a rhombus.
Square
A quadrilateral is called a square if all its sides are congruent and all its angles are
congruent. Thus a square is a parallelogram with the properties of a rectangle as
well as those of a rhombus.
Properties of a square
1) The diagonal of a square divides it into two congruent triangles.
2) The opposite sides of a square are equal.
3) The opposite angles of a square are equal.
4) The consecutive angles of a square are supplementary.
5) The diagonals of a square are equal and bisect each other at right angles.
6) The diagonals of a square bisect the opposite angles.
Example 1
If ABCD is a parallelogram and m A = 600, find m B, m L and m D.
Solution:
m B = 1200 - The consecutive angles are supplementary.
m L = 600 - The opposite angles are equal.
m D = 1200 - The opposite angles are equal.
Example 2
PQRS is a parallelogram where A is the midpoint of PQ and B is the midpoint of
SR. Prove that PABS is a parallelogram.
Solution:
Since A and B are both equidistant from seg.PS, seg.QR and seg.AB is parallel to
seg.PS. Since seg.PQ seg.SR, seg.PA seg.SB. Therefore PABS is a
parallelogram as both pairs of opposite sides are parallel.
Example 3
ABCD is a parallelogram
Find length
Solution:
,
Example 4
In a parallelogram ABCD, m A = x0, m B = ( 3x0 + 200 ), Find x, m C and
m D.
Solution:
m A + m B = 1800
x0 + 3x0 + 200 = 1800
4x0 = 1600, x = 400
m A = 400, m B = 1400
Hence m C = 400 and m D = 1400
Example 5
Which of the following statements are true ?
a) Every rectangle is a parallelogram
b) Every rhombus is a rectangle
c) Every square is a rhombus
d) Every rectangle which is a rhombus is a square.
e) Every square is a parallelogram
f) The diagonals of a parallelogram are congruent.
Solution:
a) True
b) False
c) True
d) True
e) True
f) False
CHAPTER 4: PERIMETER AND AREA
4.1 Perimeter
Perimeter is the sum of the lengths of all the sides of a figure. Thus perimeter is a
measure of lengths. In case of a circle the perimeter means its circumference.
The two dimensional space enclosed by the perimeter is called area. Since it is a
two-dimensional space, area is measured in square units like cm2 or square feet.
4.2 Square
In a square all the sides are equal in length. Therefore, if length of one side is a, the
sum of all sides or the perimeter is P = 4a.
Figure 4.1
Area of a square = a2 or the square of any one side.
4.3 Rectangle
A rectangle has 2 pairs of equal sides. Therefore if a is the length of one side and b
is the length of the other, perimeter P = a + b + a + b = 2 ( a + b ).
Figure 4.2
Area of a rectangle is the product of two consecutive sides. A = ab.
4.4 Parallelogram
In a parallelogram the lengths of the opposite sides are equal. Consider the
parallelogram ABCD (figure 4.3).
Figure 4.3
seg. AL and seg. BM are perpendiculars on the line containing CD. l (AL) is
altitude of the parallelogram.
l (AB) = a
l (AD) = b
l (AL) = h
l (BM) = h
Consider ALD and BMC. Both are right triangles.
l (AD) = l (BC) opposite sides of a parallelogram.
l (AL) = l (BM) altitudes of a parallelogram.
ALD BMC
Therefore areas of these two triangles are equal.
Consider parallelograms ABCD and ABML. Their areas are equal.
Parallelogram ABML is a rectangle. Therefore area of ABML = a h.
Thus area of a parallelogram is a product of one base and its corresponding
altitude.
A = ah
The perimeter of the parallelogram = 2 ( a + b ).
4.5 Triangle
Just as the area of parallelogram was derived from the area of rectangle, the area of
a triangle can be derived from the area of a parallelogram. Consider the triangle
PQR (figure 4.4). If a line is drawn through P parallel to RQ and another line is
drawn through Q parallel to PR they will intersect at O. POQR is a parallelogram
with PQ as its diagonal.
Figure 4.4
Recall that the diagonal of a parallelogram divides the parallelogram into two
congruent triangles.
Area of parallelogram POQR = 2 Area PQR or
Area PQR = Area of parallelogram POQR.
Area of parallelogram POQR = bh where h is the altitude on the base with length h.
Area PQR = bh
Area of a triangle is half the product of one base and the corresponding altitude.
The perimeter of the triangle is simply the sum of all its sides.
P = ( a + b + c ) in figure 4.4.
4.6 Trapezoids
The perimeter of a trapezoid is the sum of all its sides.
Perimeter of LMNO = a + b1 + a1 + b
Figure 4.5
If a trapezoid MNPQ which is congruent to LMNO is drawn on seg.MN a
parallelogram LQPO is obtained ( figure 4.6).
Figure 4.6
Area of a parallelogram LQPO = 2 Area of trapezoid LMNO
( a + a1 ) h = 2 Area of trapezoid LMNO
Area of trapezoid LMNO =
Area of a trapezoid is half the product of the sum of its bases with its altitude.
4.7 Circles
Perimeter of a circle is the circumference which is given by the formula.
Perimeter of a circle = 2 r where r is the radius, = and has been discovered by
ancient Greeks, by actually dividing the circumference of a circle by its diameter .
Area of a circle = r2 where ' r ' is the radius of the circle.
Example 1
If the area of a square is 16 sqft. , find the length of each side .
Solution:
Area of a square = (length of one side)2 length of one side = 4 ft
Example 2
If the perimeter of a square = 24 inches, what is its area?
Solution:
If perimeter 24 inches, length of each side = = 6 inches.
Area = (6 inches)2 = 36 square inches.
Example 3
The perimeter of a rectangle is 36 cm, find its area if the length of one side is 12
cm.
Solution:
Area = 72 cm2
Length of the rectangle = 12 and breadth = y
2 12 + 2 y = 36 24 + 2y = 36
or 2y = 12 y = 6 cm
Area = xy = 72 cm2
Example 4
The length of a rectangle is 10 cm and its perimeter is 30 cm. Find the area of this
rectangle.
Solution:
Perimeter = 30 cm , length = 10 cm, breadth = y
2 10 + 2xy = 30
20 + 2y = 30
2y = 10
i.e. y = 5
area = 10 5
= 50
Example 5
Find the perimeter and area of the parallelogram PQRS given below :
Solution:
Perimeter = 2 ( 3.5 + 4 )
= 15 cm
Area = 4 3
= 12 cm
Example 6
Find the perimeter and area of ABC & LMN
Solution:
Perimeter ABC = 3 + 4 + 5 = 12 cm
Area of ABC = 4 3 = 6 cm2
Perimeter LMN = 36 + 36 + 52 = 124 ft.
Area of LMN = 52 25 = 650 sq.ft.
Example 7
Find the altitude of a right triangle with area 50 sq.ft. and base 8 ft.
Solution:
Area of a right triangle =
50 = altitude = 12.5 sq.ft.
Example 8
Find the area of the trapezoid ABCD.
Solution:
Area of the trapezoid ABCD =
= 12 cm2
Example 9
Find the circumference of a circle with area 25 sq. ft.
Solution:
Area of the circle = r2 = 25
r = 5 ft.
circumference = 2 r
= 10
Example 10
Find the area of a circle with circumference 30 cm.
Solution:
Circumference = 2 r = 30
r = 15 cm
area = r2
= 152
= 225
CHAPTER 5: SIMILARITY
5.1 Introduction
The concept of similarity bears close resemblance to the concept of congruence.
Congruent figures are exact replicas of each other. They have the same shape and
the same size. Now consider figures that have the same shape but not the same
size. Such figures look 'similar' but in essence are simply proportionate to each
other.
The concept of similarity has tremendous applications. In constructing a bridge or
a building or a tower first the actual design is scaled down to the size of the paper.
The actual structure and the design on paper have the same shape but not the same
size. Here we introduce the concept of ratio and proportionality.
5.2 Ratio and Proportionality
Ratio: Ratio is a comparison of two numbers expressed in the simplest fraction
form. If a city covers an area of 100 square miles and another city covers 200
square miles, the 'ratio' of their area is expressed as 100 : 200 or on simplification 1
: 2. This means that the second city is twice as large as the first.
Ratio is a very useful way of comparing two numbers. If the age of the father is 50
years and that of the son is 10 years the ratio of their ages is 50 :10 or 5 : 1 which
means that the father is five times the son’s age.
If in a linear pair the ratio of angles is 1 : 2, it is possible to find the exact measure
of both the angles. If the measure of the smaller angle is x, the measure of the
bigger angle is 2x. Therefore x : 2x = 1 : 2.
Since a linear pair of angles sum upto 1800
x + 2x = 1800
3x = 1800
x = 600
2x = 1200
The two angles are 600 and 1200
Proportionality: Compare the drawing of a bridge on a paper with the actual
structure. They look similar because the ratios of height to width to length are the
same in both the cases.
The equation which shows that two ratios are equal is called proportion. The
design on paper and the actual structure look the same because they are
proportionate to each other.
Equality in ratios is expressed as follows:
The number at the end i.e. 3 and 10 are called extremes and the numbers in the
middle are called means.
Proportions have four properties.
1) Cross Product Property
If
This is also called the cross multiplication property.
2) Switching or exchange property.
If
3) Upside down or inverting property.
If
4) Denominator addition or subtraction property.
Example 1
A segment measuring 10 cm is divided into two parts in the ratio 1 : 3. What is the
length of each part?
Solution:
Let the length of one part of the segment be x then that of the other will be 3x .
Given that x + 3x = 10 cm.
or 4x = 10 cm.
x = 2.5 cm.
Therefore one segment measures 2.5 cm. and the other 7.5 cm.
Example 2
If the number of apples in a bag is 12 and the number of peaches is 3, what is the
ratio of apples to peaches ?
Solution:
Number of apples = 12
Number of peaches = 3
Ratio of apples to peaches is 12 : 3 or 4 : 1.
Example 3
Two complementary angles are in the ratio 1 : 2 what is their measure ?
Solution:
Let one angle be x and the other 2x
x + 2x = 900
3x = 900
x = 300
2 x = 600
The two angles measure 300 and 600.
Example 4
A 500 ft tall building is drawn as 25 cm tall on a paper. If its width is drawn as 2
cm what is the actual width of the building ?
Solution:
5.3 Similar Polygons
Polygons are said to be similar if :
a) there exists a one to one correspondence between their sides and angles.
b) the corresponding angles are congruent and
c) their corresponding sides are proportional in lengths.
Consider the polygons ABCD and LMNO in the figure 5.1.
Figure 5.1
Their corresponding angles are equal but their sides are not proportional. Hence
they are not similar. Now the sides may be proportional but the angles may not be
congruent. For instance we have polygons like PQRS and HIJK (figure 5.2)
Figure 5.2
Again they are not similar. Thus to be similar polygons must satisfy both, the
condition of congruent angles and that of proportionate sides. Figure 5.3 shows
some similar polygons.
Figure 5.3
5.4 Basic Proportionality Theorem
If a line is drawn parallel to one side of a triangle and it intersects the other two
sides at two distinct points then it divides the two sides in the same ratio.
Figure 5.4
Figure 5.4 shows triangle PQR with line l paralled to seg.QR. l intersects seg.PQ
and seg.PR at S and T respectively.
To prove that
Join S to R and Q to T
Consider PTS and QTS
Areas of triangles with same height are in the ratio of their bases.
Similarly
But A ( QTS ) = A ( SRT ) as they have a common base seg.ST and their
heights are same as they are between parallel lines.
Thus the line l which is parallel to seg.QR divides seg.PQ and seg.PR in the same
ratio.
5.5 Angle bisector theorem
In a triangle the angle bisector divides the opposite side in the ratio of the
remaining sides. This means that for a ABC ( figure 5.5) the bisector of A
divides BC in the ratio .
Figure 5.5
To prove that
Through C draw a line parallel to seg.AD and extend seg.BA to meet it at E.
seg.CE seg.DA
BAD AEC , corresponding angles
DAC ACE , alternate angles
But BAD = DAC , given AEC ACE
Hence AEC is an isosceles triangle.
seg.AC seg.AE
In BCE AD CE
Thus the bisector divides the opposite side in the ratio of the remaining two sides.
Example 1
In the diagram seg.PQ seg.BC
l (seg.AP) = 30 ft.
l (seg.PB) = 20 ft.
l (seg.QC) = 16 ft.
Find l (seg.AC).
Solution:
Example 2
In trapezium ABCD, seg.AB seg.DC. PQ DC
l (seg.AP) = 8 l (seg.PD) = 10
l (seg.BQ) = 6 find l seg.QC.
Solution:
Example 3
If in a triangle PQR a line parallel to QR cuts PQ and PR at x and y respectively,
such that l (seg.PX) = 12 , l (seg.XQ) = 8 and l (seg.PY) = 9 find (seg.YR).
Solution:
Example 4
In ABC, seg.BP is the bisecter of B. If
l (seg.AB) = 3, l (seg.BC) = 5 and
l (seg.AP) = 1.5, find l (seg.PC).
Solution:
5.6 Similar Triangles
Two triangles are similar if their corresponding angles are congruent and their
corresponding sides are proportional. It is however not essential to prove all 3
angles of one triangle congruent to the other, or for that matter all three sides
proportional to the other. Out of these if some particular conditions get satisfied the
rest automatically get satisfied. Those particular conditions would be sufficient to
ensure similarity. A group of sufficient conditions is called as a test for similarity.
These tests are based on two basic principles.
1) In 2 triangles if the corresponding angles are congruent, their corresponding
sides are equal.
2) If the sides of 2 triangles are proportional then the corresponding angles are
congruent.
A - A Test: If two angles of one triangle are congruent with the corresponding two
angles of another triangle, the two triangles are similar.
The sum of all three angles of a triangle is 1800. Therefore if two angles are
congruent the third is automatically congruent. Therefore the sufficient condition
requires only two angles to be congruent.
S A S Test: If two sides of one triangle are proportionate to the two corresponding
sides of the second triangle and the angles between the two sides of each triangle
are equal the two triangles are similar.
S S S Test: If the three sides of one triangle are proportional to the three
corresponding sides of other triangles, then two triangles are similar.
Example 1
ABC and DEF have a one to one correspondence such that
l (seg.AB) = 6 l (seg.DE) = 3
l (seg.AC) = 8 l (seg.DF) = 4
l (seg.BC) = 10 l (seg.EF) = 5. Are the two triangles similar ? If so, why ?
Solution:
Yes. The two triangles are similar because their corresponding sides are
proportionate.
Example 2
If ABC DEF PQR such that l (seg.AB) = 7 l (seg.DE) = 8 l (seg.PQ) = 4
and l (seg.EF) = 6, find l (seg.BC) and l (seg.QR).
Solution:
If ABC DEF PQR
Example 3
a)
b)
c)
Each pair of triangles is similar. By which test can they be proved to be similar ?
Solution:
a) S A S test
b) A A test
c) S S S test
5.7 Properties of Similar triangles
Perimeters of similar triangles: Perimeters of similar triangles are in the same
ratio as their corresponding sides and this ratio is called the scale factor.
In figure 5.6 there are two similar triangles . LMN and PQR.
Figure 5.6
This ratio is called the scale factor.
Perimeter of LMN = 8 + 7 + 10 = 25
Perimeter of PQR = 6 + 5.25 + 7.5 = 18.75
Thus, the perimeters of two similar triangles are in the ratio of their scale factor.
Areas of similar triangles: The ratio of the areas of two similar triangles is equal
to the ratio of the squares of the corresponding sides, i.e. the square of the scale
factor.
Figure 5.7
D ABC PQR
To prove that
Draw perpendicular from A and P to meet seg.BC and seg.QR at D and S
respectively.
Since ABC ~ PQR
also B Q
In ABD and PQS also B Q and ADB PSQ
ABD PQS by A A test.
Thus the areas of two similar triangles are in the same ratio as the square of their
scale factors.
Example 1
Areas of two similar triangles are 144 sq.cm. and 81 sq.cm. If one side of the first
triangle is 6 cm then find the corresponding side of the second triangle.
Solution :
= 4.5 c.m.
Example 2
The side of an equilateral triangle ABC is 5 cm. Find the length of the side of
another equilateral PQR whose area is four times area of ABC.
Solution:
Since both triangles are equilateral they are similar.
Example 3
The corresponding sides of two similar triangles are 4 cm and 6 cm. Find the ratio
of the areas of the triangles.
Solution:
Example 4
ABC PQR such that l (seg.AB) : l (seg.PQ) that is 8 : 6. If area of ABC is 48
sq.cm what is the area of the smaller triangle.
Solution:
Example 5
In a trapezium ABCD, side AB CD. The diagonals AC and BD cut each other at
M.
Prove that
Solution:
To Prove that
Consider AMB and CMD.
AMB CMD, vertical angles
BAM DCM alternate angles
By AA test AMB ~ CMD \
CHAPTER 6:
THEOREM OF PYTHAGORAS AND THE RIGHT TRIANGLE
6.1 The Right Triangle
Figure 6.1
ABC is a right triangle, hence m ABC = 900. Therefore m A and m C are
complementary ( figure 6.1). Now seg.BD is a perpendicular onto seg.AC (figure
6.2).
Figure 6.2
Seg.BD divides ABC into two right triangles BDC and ADB ( figure 6.2). It
can be easily proven that these two triangles are similar to the parent ABC and
therefore similar to each other.
Proof :
Consider ABC and BDC
ABC BDC right angles and
BCA DCB same angle
by AA test ABC BDC (1)
Similarly consider ABC and ADB.
ABC ADB right angle
CAB DAB same angle
by AA test ABC ADB (2)
from (1) and (2) ABC ADB BDC.
Since ABC BDC
and ABC ADB
In (A) l (BC) is repeated and in (B) l (seg.AB) is repeated at the means. This is
referred to as the geometric mean.
The two proportions (A) and (B) obtained by the similarity of ADB and BDC
with the original triangle are stated as a theorem as follows:
If an altitude seg.BD is drawn to the hypotenuse seg.AC of a right triangle ABC
then each leg , i.e. seg.AB and seg.BC is the geometric mean between the
hypotenuse and seg.DA and seg.DC respectively ( refer figure 6.2).
The similarity of BDC and ADB gives the proportion.
i.e. seg.BD is the geometric mean between seg.AD and seg.CD. The altitude drawn
on the hypotenuse is the geometric mean between the two segments the hypotenuse
is cut into.
Example 1
Find the geometric mean between :
a) 2 and 18
b) 4 and 16
c) 9 and 25
Solution :
a)
b)
c)
Example 2
Find x.
Solution :
The square of the altitude to the hypotenuse is equal to the product of the segments
cut on the hypotenuse.
x2 = 12 3
x2 = 36
x = 6.
Example 3
Find y.
Solution :
6.2 The Theorem of Pythagoras
Figure 6.3
ABC is a right triangle.
l (AB) = c
l (BC) = a
l (CA) = b
CD is perpendicular to AB such that
ABC CBD
or l (BC)2 = l (AB) l (CD)
a2 = c x = cx (1)
ABC ACD
or l (AC)2 = l (AB) l (AD)
b2 = c y = cy (2)
Therefore, from (1) and (2)
a2 + b2 = cx + cy
= c ( x + y )
= c c
= c2
a2 + b2 = c2
The square of the hypotenuse is equal to the sum of the squares of the legs.
Converse of Pythagoras Theorem : In a triangle if the square of the longest side
is equal to the sum of the squares of the remaining two sides then the longest side
is the hypotenuse and the angle opposite to it, is a right angle.
Figure 6.4
Given that in ABC l (AC)2 = l (AB)2 + l (BC)2
To prove that ABC is a right triangle or that
m ABC = 900.
Construct a PQR such that
l (QR) = l (BC) , l (PQ) = l (AB) & PQR = 900.
In PQR l (PR)2 = l (PQ) 2 + l (QR)2
= l (AB)2 + l (BC)2
But l (AB)2 + l (BC)2 = l (AC)2
l (PR)2 = l (AC)2
or that l (PR) = l (AC) (1)
In ABC and PQR
l (AB) = l (PQ) construction
l (BC) = l (QR)construction
l (AC) = l (PR) from (1)
ABC PQR by S S S.
M B = M Q
Since m Q = 900 B is a right angle and ABC is a right triangle. Therefore if
the square of the length of the longest side is equal to the sum of the squares of the
other two sides the triangles is a right triangle.
Example 1
The length of a rectangle is 4 ft. and the breadth is 3ft. What is the length of its
diagonal?
Solution :
ABCD is a rectangle such that l (seg.AB) = 4 , l (seg.BC) = 3. m ABC is 900.
ABC is a right triangle.
l (AC) 2 = l (AB)2 + l (BC)2
= (4)2 + (3)2
= 16 + 9
= 25
l (AC) = 5 ft.
The length of the diagonal is 5 ft.
Example 2
A man drives south along a straight road for 17 miles. Then turns west at right
angles and drives for 24 miles where he turns north and continue driving for 10
miles before coming to a halt. What is the straight distance from his starting point
to his terminal point ?
Solution :
The man drove from A C D E
EB is perpendicular to AC .
l (AB) = 17 - 10 = 7 miles
l (EB) = 24 miles
ABE is a right triangle
l (AE)2 = l (AB)2 + l (EB)2
= (7)2 + (24)2
= 49 + 576
= 625
l (AE) = 25 miles.
Example 3
ABC is a right triangle. m ACB = 900 , seg.AQ bisects seg.BC at Q.
Prove that 4 l (AQ)2 = 4 l ( AC)2 + BC2
Solution :
ABC is a right triangle
l (AQ)2 = l (AC)2 + l (CQ)2
or 4 l (AQ)2 = 4 l (AC)2 + 4 l (CQ)2
= 4 l (AC)2 + { 2 l (CQ) }2
2 l (CQ) = l (BC) as Q is the midpoint of BC
l (AQ)2 l (AC)2 + l (BC)2.
In any triangle if the square of the longest side is greater than the sum of the
squares of the other two sides the triangle is an obtuse triangle. If however the
square of the longest side is less than the sum of the squares of the other two sides
the triangle is an acute triangle. Given the lengths of the three sides of a triangle
are a, b and c where c > a and b. If c2 > a2 + b2 ABC is an obtuse and if
c2 < a2 + b2 ABC is an acute triangle.
6.3 Special Right Triangles
The 300 - 600 - 900 triangle: If the angles of a triangle are 300 , 600 and 900 then
the side opposite to 300 is half the hypotenuse and the side opposite to 600 is
times the hypotenuse.
Figure 6.5
In ABC A = 900 B = 600 and C = 300
To prove that l (AB) = l (BC) and l (AC) =
Take a point D on ray BA such that seg.AD seg.AB and join CD.
In ABC and ACD
AB AD construction
CAB CAD both are right angles
AC AC common side
BAC DAC (SAS)
ACB ACD corresponding angles of congruent triangles are equal.
but m ACB = 300
m DCB = 600
DCB is an equilateral triangle
l (seg.DC) = l (seg.BC) = l (seg.DB) (1)
but l (seg. AB) = l (DB) (2)
(3) from (1) and (2).
In right triangle ABC
l (seg.AB)2 + l (seg.AC)2 = l (seg.BC)2 Pythagoras
{ l (seg.BC)2 } + l (seg.AC)2 = l (seg.BC)2from (3)
l (seg.AC)2 = l (seg.BC)2 - l (seg.BC)2
= l (BC)2
The 450 - 450 - 900 triangle: If the angles of a triangle are 450 - 450 - 900 then the
perpendicular sides are times the hypotenuse.
In ABC A = 450 , B = 900 and C = 450.
Figure 6.6
To prove that AB = BC =
By Pythagoras theorem
l (seg.AB)2 + l (seg.BC)2 = l (seg.AC)2
l (seg.AB) = l (seg.BC) . ABC is isosceles.
l (seg.AB)2 + l (seg.BC)2 = 2 l (seg.AB)2 = l (seg.AC)2
Example 1
LMNO is a parallelogram such that m LON = 300 and l (seg.LO) = 12 cm. If
seg.MP is the perpendicular distance between seg.LM and seg.ON ,
find l (seg.MP).
Solution :
l (seg.MP) = 6
Since m LON = m MNP = 300
MNP is a 300 - 600 - 900 triangle
l (seg.MP) = l (seg.MN)
l (seg.MN) = l (seg.LO) = 12 cm
l seg.MP = 12 cm
= 6 cm.
Example 2
PQR is an acute triangle seg.PS is perpendicular to seg.QR and seg.PT bisects
QR.
Prove that l (seg.PR)2 + l (seg.PQ)2 = l (seg.PT)2 + l (seg.QT)2
Solution :
To prove that l (seg.PR)2 + l (seg.PQ)2 = l (seg.PT)2 + l (seg.QT)2
In PRS, by Pythagoras theorem
l (seg.PR)2 = l (seg.PS)2 + l (seg.SR)2
= l (seg.PS)2 + [ l (seg.ST) + l (seg.TR) ]2
= l (seg.PS)2 + l (seg.ST)2+ 2 l (seg.ST )
l (seg.TR) + l (seg.TR) 2 (1)
In PQS, by Pythagoras theorem
l (seg.PQ)2 = l (seg.PS)2 + l (seg.QS)2
= l (seg.PS)2 + [ l (seg.QT) + l (seg.ST) ]2
= l (seg.PS)2 + l (seg.QT)2 + 2 l (seg.QT)
l (seg.ST) + l (seg.ST)2 (2)
From (1) and (2)
l (seg.PR)2 + l (seg.PQ)2
= l (seg.PS)2 + l (seg.ST)2 + 2 l (seg.ST) l
(seg.TR) + l (seg.TR)2 + l (seg.PS)2
+ l (seg.QT)2 - 2 l (seg.QT)
l (seg.ST) + l (seg.ST)2.
Since l (seg.QT) = l (seg.TR)
l (seg.PR)2 + l (seg.PQ)2
= 2 l (seg.PS)2 + 2 l (seg.ST)2 + l (seg.QT)2 + 2 l (seg.ST)
l (seg.QT) l (seg.ST) l (seg.QT)
= 2 l (seg.PS)2 + 2 l (seg.ST)2 + 2 l (seg.QT)2
= 2 { l (seg.PS)2 + l (seg.ST)2 } + 2 l (seg.QT)2
= 2 l (seg.PT)2 + 2 l (seg.QT)2
CHAPTER 7 : CIRCLE
7.1 Introduction
A circle is defined as a set of all such points in a given plane which lie at a fixed
distance from a fixed point in the plane. This fixed point is called the center of the
circle and the fixed distance is called the radius of the circle (see figure 7.1).
Figure 7.1
Figure 7.1 shows a circle where point P is the center of the circle and segment PQ
is known as the radius. The radius is the distance between all points on the circle
and P. It follows that if a point R exists such that l (seg.PQ) > l (seg.PR) the R is
inside the circle. On the other hand for a point T if l (seg.PT) > l (seg.PQ) T lies
outside the circle. In figure 7.1 since l (seg.PS) = l (seg.PQ) it can be said that point
S lies on the circle.
7.2 Lines of a Circle
The lines in the plane of the circle are classified into three categories (figure 8.2).
a) Lines like l which do not intersect the circle.
b) Lines like m which intersect the circle at only one point.
c) Lines like n which intersect the circle at two points..
Figure 8.2
Lines like m are called tangents. A tangent is a line that has one of its points on a
circle and the rest outside the circle.
Line n is called a secant of the circle. A secant is defined as any line that intersects
a circle in two distinct points.
A segment whose end points lie on a circle is called a Chord . In figure 8.2 AB is a
chord of the circle. Thus a chord is always a part of secant. A circle can have an
infinite number of chords of different lengths (figure 7.3)
Figure 7.3
The longest chord of the circle passes through its center and is called as
the diameter. In figure 7.3 chord CD is the diameter. It can be noticed
immediately that the diameter is twice the radius of the circle. The center of the
circle is the mid point of the diameter. A circle has infinite diameters and all have
the same length.
Example 1
A, B, C & D lie on a circle with center P. Classify the following segments as radii
and chords.
PA, AB, AC, BP, DP, DA, PC, BC, BD, CD.
Solution:
Example 2
Name the secant and the tangent in the following figure :
Solution:
secant - l
tangent - m
Example 3
P is the center of a circle with radius 5 cm. Find the length of the longest chord of
the circle.
Solution:
10 cm.
If r = 5 cm d = 2r = 10 cm.
Diameter is the longest chord.
Therefore length of longest chord = 10 cm.
7.3 Arcs
The angle described by any two radii of a circle is called the central angle. Its
vertex is the center of the circle. In figure 8.4 APB is a central angle. The part of
the circle that is cut by the arms of the central angle is called an arc. AB is an arc
and so is AOB . They are represented as & .
Figure 7.4
is called the minor arc and is the major arc. The minor arc is always
represented by using the two end points of the arc on the circle. However it is
customary to denote the major arc using three points. The two end points of the
major arc and a third point also on the arc. If a circle is cut into two arcs such that
there is no minor or major arc but both the arcs are equal then each arc is called
a semicircle.
An arc is measured as an angle in degrees and also in units of length. The measure
of the angle of an arc is its central angle and the length of the arc is the length of
the portion of the circumference that it describes.
angle of an arc AB = m
length of an arc AB = l
Since the measure of the angle of an arc is its central angle, if two central angles
have equal measure then the corresponding minor arcs are equal.
Conversely if two minor arcs have equal measure then their corresponding central
angles are equal.
7.4 Inscribed angles
Whereas central angles are formed by radii, inscribed angles are formed by chords.
As shown in figure 8.5 the vertex o of the inscribed angle AOB is on the circle.
The minor arc cut on the circle by an inscribed angle is called as the
intercepted arc.
Figure 7.5
Theorem: The measure of an inscribed angle is half the measure of its intercepted
arc.
Proof: For a circle with center O BAC is the inscribed angle and arc BXC is the
intercepted arc. To prove that m BAC = 1/2 m (arc BXC). There arise three cases as
shown in figure 8.6 (a), 8.6 (b) and 8.6 (c).
Click here to enlarge
Figure 7.6 (a) Figure 7.6 (b) Figure 7.6 (c)
Case 1: The center is on the angle figure 7.6 (a) join c to O.
OAC is an isosceles triangle as seg.OA = seg.OC.
Assume m OAC = m OCA = P
m COA = 180 - 2 P as sum of all the angles of a triangle is 1800.
COA & COB form a linear pair. Therefore they are supplementary.
m COB = 1800 - m COA
= 1800 ( 1800 - 2P)
= 2P
= 2 m BAC
But m BAC = m (arc BXC)
m BAC = m (arc BXC)
Case 2: Figure 7.6 (b). The center is in the interior of the angle. in this case let D
be the other end point of the diameter drawn through A.
Let arc CMD be intercepted by CAD and let arc BND be that which is intercepted
by DAB.
From case 1 m CAD = m ( arc CMD)
and m DAB = m (arc BND)
m CAD + m DAB = m (arc CMD) + m (arc BND)
m BAC = { m ( arc CMD) + m (arc BND) }
= m (arc BXC)
Case 3: The center is in the exterior of the angle. Again let D be the other end
point of the diameter drawn through A. Let arc CMD be the one intercepted
by CAD and let arc BND be the one intercepted by DAB.
From case 1 m CAD = m (arc CMD)
m DAB = m (arc BND)
m CAD - m DAB = m (arc CMD) - m (arc BND)
m CAB = { M (arc CMD) - m (arc BND) }
= m (arc BXC)
Thus it is proved that the measure of the inscribed angle is half that of the
intercepted arc.
Theorem: If two inscribed angles intercept the same arc or arcs of equal measure
then the inscribed angles have equal measure.
Figure 7.7
In figure 7.7 CAB and CDB intercept the same arc CXB.
Prove that m CAB = CDB.
From the previous theorem it is known that
m CAB = m (arc CXB) and also
m CDB = m (arc CXB)
m CAB = m CDB
Therefore if two inscribed angles intercept the same arc or arcs of equal measure
the two inscribed angles are equal in measure.
Theorem: If the inscribed angle intercepts a semicircle the inscribed angle
measures 900.
Figure 8.8
The inscribed angle ACB intercepts a semicircle arc AXB (figure 8.8). We have to
prove that m ACB = 900.
m ACB = m (arc AXB)
= (1800)
= 900
Therefore if an inscribed angle intercepts a semicircle the inscribed angle is a right
angle.
Example 1
a) In the above figure name the central angle of arc AB.
b) In the above figure what is the measure of arc AB.
c) Name the major arc in the above figure.
Solution:
a) AOB
b) 800. The measure of an arc is the measure of its central angle.
c) Arc AXB
Example 2
a) In the above figure name the inscribed angle and the intercepted arc.
b) What is m (arc PQ)
Solution:
a) inscribed angle - PRQ
intercepted arc - arc PQ
b) 600. The measure of an intercepted arc is twice the measure of its inscribed
angle.
Example 3
Ð PAQ and PBQ intercept the same arc PQ what is the m PBQ and m (arc PQ) ?
Solution:
m PBQ = 400 If two inscribed angles intercept the same arc their measures are equal m (arc
PQ) = 800 as m (arc) = 2m (inscribed angle).
7.5 Some properties of tangents, secants and chords
Theorem: If the tangent to a circle and the radius of the circle intersect they do so
at right angles:
Figure 7.9 (a) Figure 7.9 (b)
In figure 7.9 (a) l is a tangent to the circle at A and PA is the radius.
To prove that PA is perpendicular to l , assume that it is not.
Now, with reference to figure 7.9 (b) drop a perpendicular from P onto l at say B.
Let D be a point on l such that B is the midpoint of AD.
In figure 7.9 (b) consider PDB and PAB
seg.BD seg.BA ( B is the midpoint of AD)
PBD PBA ( PB is perpendicular to l ) and
seg.PB = seg.PB (same segment)
PBD PAB (SAS)
seg.PD seg.PA corresponding sides of congruent triangles are congruent.
D is definitely a point on the circle because l (seg.PD) = radius.
D is also on l which is the tangent. Thus l intersects the circle at two distinct points A
and D. This contradicts the definition of a tangent.
Hence the assumption that PA is not perpendicular to l is false. Therefore PA is
perpendicular to l.
Angles formed by intersecting chords, tangent and chord and two secants: If
two chords intersect in a circle, the angle they form is half the sum of the
intercepted arcs.
In the figure 7.10 two chords AB and CD intersect at E to form 1 and 2.
Figure 7.10
m 1 = (m seg.AD + m seg.BC) and
m 2 = (m seg.BD + m seg.AC)
Tangent Secant Theorem: If a chord intersects the tangent at the point of
tangency, the angle it forms is half the measure of the intercepted arc. In the figure
7.11, l is tangent to the circle. Seg.AB which is a chord, intersects it at B which is
the point of tangency.
Figure 7.11
The angles formed ABX and ABY are half the measures of the arcs they
intercept.
m 1 = m (arc ACB)
m 2 = m (arc AB)
This can be proved by considering the three following cases.
Figure 7.12
O is the center of the circle
Line DBC is tangent to it at B.
BA is the chord in question. X is a point on the circle on the C side of BA and Y is
on the D side by BA.
Arc AXB can be
a) Semi circle
b) minor arc
c) major arc
Case 1 : Assume arc AXB is a semicircle when ABC intercepts a semicircle the
chord AB passes through the center. Therefore m ABC = 900 (a tangent is always
perpendicular to the diameter that intersects it at the point of tangency).
m (arc BXA) = 1800 (arc BXA is a semi circle)
m (arc BXA) = 1800 = 900
m ABC = m (arc BXA)
Case 2 : Assume that ABC intercepts a minor arc. Therefore as seen in figure 8.13 the
center O lies in the exterior of ABC.
Figure 7.13
m ABC = 900 - m ABO
m ABO = 900m ABC ----------- (1)
But m ABO = m OAB
(as OAB is an isosceles triangle )
m OAB = 900 - m ABC ----------- (2)
(1) + (2)
m ABO + m OAB = 180 - 2 m ABC
Since m ABO + m OAB = 180 - m BOA
180 - m BOA = 180 - 2 m ABC
i.e. m BOA = 2 m ABC
m ABC = m BOA
m ABC = m ( arc AXB )
Case 3 :
Figure 7.14
If ABC intercepts a major arc, the center of the circle O will lie in the interior asABC .
See figure 7.14.
Now ADB intercepts a minor arc AYB.
m AOB = m (arc AYB)
1800 - m ADB = { 3600 - m (arc AXB) }
1800 - m ADB = 1800 - m (arc AXB)
m ADB = m (arc AXB)
If two secants intersect outside a circle half the difference in the measures of the
intercepted arcs gives the angle formed by the two secants.
In figure 7.15 l and m are secants. l and m intersect at O outside the circle. The
intercepted arcs are and .
COD = ( m - m )
Figure 7.15
Conclusion :
(a) If two chords intersect in a circle the angle formed is half the sum of the
measures of the intercepted arcs.
(b) Angle formed by a tangent and a chord intersecting at the point of tangency is
half the measure of the intercepted arcs.
(c) Angle formed by two secants intersecting outside the circle is half the
difference of the measures of the intercepted arcs.
Example 1
In the above figure seg.AB and seg.CD are two chords intersecting at X such that
m AXD = 1150 and m (arc CB) = 450 . Find m arc APD.
Solution:
m arc APD = 1850
m AXD = { m (arc APD) + m (arc CB) }
m ( arc APD) = 2 m AXD - m (arc CB)
= 2 1150 450
= 1850
Example 2
l is a tangent to the circle at B. Seg. AB is a chord such that m ABC = 500. Find the m
(arc AB).
Solution:
m (arc AB) = 1000
m ABC = m (arc AB)
50 = m (arc AB)
m (arc AB) = 1000
Example 3
l and m are secants to the circle intersecting each other at A. The intercepted arcs are arc PQ
and arc RS if m PAQ = 250 and m ROS = 800 find m (arc PQ).
Solution:
m ( arc PQ) = 300
m PAQ = { m (arc RS) - m (arc PQ)
2 m PAQ = m (arc RS) - m (arc PQ)
\ m (arc PQ) = m (arc RS) - 2 m PAQ
= 800 - 500
= 300
7.6 Chords and their arcs
Theorem: If in any circle two chords are equal in length then the measures of their
corresponding minor arcs are same.
As shown in figure 8.16 AB and CD are congruent chords. Therefore according to
the theorem stated above m (arc AB) = m (arc CD) or m AOB = m COD.
Figure 7.16
To prove this, join A, B, C & D withO.
Consider AOB and COD
seg.AO seg.CO and seg.OB seg.OD (radii of a circle are always congruent).
seg.AB seg.CD (given)
AOB COD ( S S S )
AOB COD (corresponding angles of congruent triangles are congruent).
Hence arce AB arc CD.
Conversely it can also be proved that in the same circle or congruent circles,
congruent arcs have their chords congruent.
Figure 7.17
In figure 7.17 if arc LM arc PQ then seg. LM seg. PQ.
To prove this, join L, M, P & Q to O.
Consider LOM and POQ
seg. OL seg. OP and seg. MO seg QO ( as all are radii ).
LOM POQ (given )
LOM POQ (SAS)
seg. LM seg PQ (corresponding sides of congruent triangles are congruent).
Theorem: The perpendicular from the center of a circle to a chord of the circle
bisects the chord.
Figure 7.18
In figure 7.18, XY is the chord of a circle with center O. Seg.OP is the
perpendicular from the center to the chord. According to the theorem given above
seg XP = seg. YP.
To prove this, join OX and OY
Consider OXP and OXY
Both are right triangles.
hypotenuse seg. OX hypotenuse seg. OY
( both are radii of the circle )
seg. OP seg OP (same side)
OXP OYP (H.S.)
seg XP seg. YP (corresponding sides of congruent triangles are congruent).
P is the midpoint of seg.XY.
Hence seg.OP which is the perpendicular from the center to the chord seg.XY
bisects the chord seg.XY.
Now consider two chords of equal length in the same circle. Their distance from
the center of the circle is same.
In figure 7.19 seg. PQ and seg. RS are two chords in the circle with center O such
thatl (seg.PQ) = l (seg.RS).
seg.OX and seg. OY are the perpendicular distances from O to seg.PQ and seg.RS
respectively.
To prove that l (seg.OX) = l (seg.OY) join O to P and R.
Since perpendicular from the center to the chord bisects the chord l ( seg.PX)
= l(seg.RY).
Now consider POX and ROY
seg. PX seg.RY
seg.OP seg.OR. Both being radii onto the circle.
POX ROY (by H.S.)
seg.OX seg.OY (corresponding sides of congruent triangles are congruent).
Figure 7.19
Thus if two chords are equal in measure they are equidistant from the center of the
circle.
The converse of this theorem is that if two chords are equidistant from the center of
the circle, they are equal in measure.
As shown in figure 7.20 if seg. HI and seg. JK are two chords equidistant from the
center of the circle, they are equal in length.
Figure 7.20
To prove that seg.HI seg.JK join OI and OK.
Consider OIP and OKQ, ( both are right triangles) .
seg.OI seg.OK, ( both are radii of the same circle).
seg.OP seg.OQ (given that chords are equidistant from the center O).
OIP OKQ (H.S.)
seg.PI seg.QK (corresponding sides of congruent triangles are congruent).
Also it is known that the perpendicular from the center bisects the chord.
Therefore, seg. HI seg JK.
Example 1
AB and CD are chords in a circle with center O. l (seg.AB ) = l (seg.CD) = 3.5 cm and
m COD = 950. Find m arc AB.
Solution:
950
m arc AB = m AOB
Since AOB COD by SSS m AOB = m COD.
Example 2
PQ is a chord of a circle with center O. Seg.OR is a radius intersecting PQ at right
angles at point T. If l (PT) = 1.5 cm and m arc PQ = 800, find l (PQ) and m arc PR.
Solution:
l (PQ) = 3
m (arc PR) = 400
Seg.OT is perpendicular to PQ and therefore bisects PQ at T.
l (.PQ) = 2 l (PT)
Seg.OR bisects arc PQ. m (arc PR) = m (arc PQ)
Example 3
Seg HI and seg. JK are chords of equal measure in a circle with center O. If the
distance between O and seg. HI is 10 cm find the length of the perpendicular from
O onto seg.JK.
Solution:
10 cm.
Chords of equal measure are equidistant from the center.
7.7 Segments of chords secants and tangents
Theorem : If two chords, seg.AB and seg.CD intersect inside or outside a circle at
P then l (seg. PA) l (seg. PB) = l (seg. PC) l (seg. PD)
Click here to enlarge
Figure 7.21 (a) Figure 7.21 (b)
In figure 7.21 (a) P is in the interior of the circle. Join AC and BD and
consider APC and BDP.
m APC = m BPD (vertical angles).
m CAP = m BDP (angles inscribed in the same arc).
APC BPD ( A A test )
( corresponding sides of similar triangles).
l (seg. PA) l (seg. PB) = l (seg. PC) l (seg. PD).
Now consider figure 7.21 (b).
P is in the exterior of the circle. Join A to C and B to D.
Consider PAC and PBD
m APC = m BPD (same angle).
m CAP = m PDB (exterior angle property of a cyclic quadrilateral).
PAC PDB ( A A test )
(corresponding sides of similar triangles).
l (seg. PA) l (seg. PB) = l (seg. PC) l (seg. PD).
Consider a secant PAB to a circle, (figure 7.22) intersecting the circle at A and B
and line PT is a tangent then l (seg. PA) l (seg. PB) = l (seg. PT)2.
Figure 7.22
P is a point in the exterior of the circle. A secant passes through P and intersects
the circle at points A & B. Tangent through P touches the circle in point T.
To prove that l (seg. PA) l (seg. PB) = l (seg. PT)2
Consider PTA and PTB.
m TPA = TPB ( same angle)
According to Tangent Secant theorem,
m ATP = m (arc AT)
= m PTB ( inscribed angle )
PTA PTB ( A A test )
(corresponding sides of similar triangles).
l (seg. PA) l (seg. PB) = l (seg. PT)2.
Theorem: The lengths of two tangent segments from an external point to a circle
are equal.
As shown in figure 7.23 seg. QR and seg. QS are two tangents on a circle with P as
its center.
Click here to enlarge
Figure 7.23
To prove that l (seg.QR) = l (seg.QS) join P to Q and R to S.
m PRQ = m PSQ = 900.
The radius and the tangent form a right angle at the point of tangency,
PRQ and PSQ are right triangles such that
seg. PR seg PS (radii of the same circle).
seg. PQ seg. PQ (same side).
PRQ PSQ (H.S)
seg.QR seg.QS (corresponding sides of congruent triangles are congruent).
l (seg.QR) = l (seg.QS).
Example 1
Two chords seg AB and seg. CD intersect in the circle at P. Given that l (seg.PC)
= l(seg.PB) = 1.5 cm and l (seg.PD) = 3 cm. Find l (seg.AP).
Solution:
l (seg AP) l (seg PB) = l (seg DP) l (seg PC)
l (seg AP) 1.5 = 3 1.5
l (seg AP) = 3 cm.
Example 2
Seg.PA and seg.PC are two secants where l (seg.PB) = 3.5 cm, l (seg.PD) = 4 cm
andl (seg.DC) = 3 cm. Find l (seg.AB).
Solution:
l (seg AP) l (seg BP) = l (seg CP) l (seg DP)
x 3.5 = 7 4
x = 8
l (seg AB) = l (seg AP) l (seg BP)
= 8 - 3.5
= 4.5 cm.
Example 3
PT is a tangent intersecting the secant through AB at P. Given l (seg. PA) = 2.5 cm.
and l (seg.AB) = 4.5 cm., find l (seg PT).
Solution:
l (seg PT)2 = l (seg PA) l (seg PB)
= 2.5 7
= 17.5
l (seg. PT) = 4.2 cm.
7.8 Lengths of arcs and areas of sectors
An arc is a part of the circumference of the circle; a part proportional to the central
angle.
If 3600 corresponds to the full circumference. i.e. 2 r then for a central angle of x0(figure
7.24) the corresponding arc length will be l such that
Figure 7.24
Analogically consider the area of a sector. This too is proportional to the central
angle. 3600 corresponds to area of the circle r2. Therefore for a central angle m0 the area
of the sector will be in the ratio:
Example 1
In a circle with the radius of 2 cm, the central angle for an arc AB is 750.
Find l(seg.AB). Also find the area of the sector AOB having a central angle of 750
Solution:
l (seg AB) = 2.6
CHAPTER 8 : SURFACE AREA AND VOLUME
8.1 Introduction to Solid Geometry
All the geometric shapes discussed in this book till now i.e. polygons, circles etc.
are planar shapes. They are called two dimensional shapes i.e. generally speaking
they have only length and breadth. In the real world however every object has
length breadth and height. Therefore they are called three dimensional objects. No
single plane can contain such objects in totality.
Consider the simplest example of a three dimensional shape - a brick.
Figure 8.1
Shown in Figure 8.1 is a brick with length 10 cm, breadth 5 cm and height 4 cm.
There cannot be a single plane which can contain the brick.
A brick has six surfaces and eight vertices. Each surface has an area which can be
calculated. The sum of the areas of all the six surfaces is called the surface area of
the brick.
Apart from surface area a brick has another measurable property. i.e. the space it
occupies. This space occupied by the brick is called its volume. Every three
dimensional (3-D) object occupies a finite volume. The 3 -D objects or geometric
solids dealt within this chapter are
(1) Prism
(2) Cube
(3) Right circular cylinder
(4) Pyramid
(5) Right circular cones and
(6) Sphere
Apart from defining these objects, methods to calculate their surface area and
volume are also incorporated in this chapter.
8.2 Prism
Any solid formed by joining the corresponding vertices of two congruent polygons
is called a prism.
Figure 8.2
The two congruent polygons are called the two bases of the prism. The lines
connecting the corresponding vertices are called lateral edges and they are parallel
to each other. The parallelogram formed by the lateral edges are called lateral
faces.
(See figure 8.2)Prisms are of two types depending on the angle made by the lateral
edges with the base.
If the lateral edges are perpendicular to the base the prism is called a right prism.
If the lateral edges are not perpendicular to the base the prism is called an oblique
prism.
Consider the right prism shown in Figure 8.3
Figure 8.3
ABCDEF is a prism ABC & FED are congruent and seg. AF, seg. CD and seg.
BE are perpendicular to the planes containing ABC & FED.
Also in ABC m ABC = 900 , l (seg. AB) = 3 cm & l (seg. AC) = 5 cm
The surface area of the prism is the sum of the surface areas of all its surfaces.
base is unknown and height is 3 cm.
8.3 The cuboid and the cube
A book, a match box, a brick are all examples of a cuboid.
The definition of a cuboid is derived from that of the prism.
A cuboid is a solid formed by joining the corresponding vertices of two congruent
rectangles such that the lateral edges are perpendicular to the planes containing the
congruent rectangles. Figure 8.3 shows a cuboid.
Figure 8.3
As can be seen in Figure 8.3 the cuboid has six surfaces and each one is congruent
and parallel to the one opposite to it.
Thus there is a pair of three rectangles which goes to make a cube.
It is already known that the area of a rectangle is the product of its length and
breadth, l b = Area.
The surface area of a cuboid is the sum of the areas of all its faces,
Now consider a cuboid made of 6 congruent squares instead of a pair of three
rectangles. This structure has l = b = h i.e. the length breadth and height are the
same. Such a cuboid is called a cube.
Cube
A cube is a square right prism with the lateral edges of the same length as that of a
side of the base. (See figure 8.4)
Figure 8.4
As can be seen in Figure 8.4 in a cube the length, breadth and height are equal.
Therefore the surface of a cube is essentially six squares of equal length.
The area of a square is the square of its length. The total surface area of the cube is
the sum of the areas of all its 6 surfaces. If length of one side of the cube is l the
area of one surface is l 2 the surface area of the cube = 6 l 2 = 6 l 2 .
To find volume of a cuboid first a unit cube has to be defined. A unit cube is a
cube with length equal to one unit.
This unit is chosen as per convenience. For small volumes 1 cm can be chosen as
the unit length . The volume of this cube is 1 cubic cm. It is represented as 1 cm3.
For larger volumes 1 meter can be used as the unit length. In this case the volume
of the unit cube is 1 cubic meter and is represented by 1m3 .
Depending on the size of the cuboid whose volume is to be measured the
appropriate unit cube is chosen.
The volume of a cuboid is the number of unit cubes that are required to fill the
cuboid completely.
Figure 8.6 shows a cuboid whose volume is to be measured. This cube has length
10 cm breadth 8 cm and height 4 cm.
The three edges OA, OB and OC are perpendicular to each other. On these sides
points LM and N are chosen such that l (seg. O l) = l (seg. ON) = unit length.
Figure 8.5
These sides seg. OL, seg. OM and seg. ON form three sides of a unit cube as
shown in Figure 8.6a Let another unit cube be placed on the right side of this unit
cube such that one of its edges lies along seg. LA and one of its face covers the
right side of the first unit cube. (See Figure 8.6b)
Figure 8.6a Figure 8.6b
Since length of seg. OA = 10 and l (seg. OL) = 1, 10 unit cubes can be arranged
along OA. Similarly l (seg. OB) = 5. 5 unit cubes can be arranged along seg.
OB. Thus to cover the face AOBD, 10 5 = 50 unit cubes will be required. They
will form the first layer towards filling the cuboid.
Since l (seg. OC) = 6 it is obvious that 6 such layers are required to fill the cuboid.
Number of unit cubes required = 50 6
= 300
Thus the space occupied by the cuboid or its volume
= 300 cubic units
= 300 unit3
In general it may be stated that for a cuboid with length l , breadth b and height h
the volume = l b h cubic units.
For a cuboid,
Surface area = 2 (l b + b h + l h)
Volume = (l b h) cubic units.
Since a cube has l = b = c
its surface area = 6 l 2 & Volume = l 3
Example 1
Draw a right hexagonal prism with height 5 cm and length of one side of the
hexagon = 2.5 cm.
Solution:
Example 2
When is a cuboid a cube ?
Solution:
If the length, breadth and height of a cuboid are equal it is a cube.
Example 3
What is the difference between a right prism and an oblique prism ?
Solution:
In a right prism, the angle formed by the lateral edges with the planes containing
the bases is 900. In oblique prism the same angle is not 900 . In oblique prism the
same angle is not 900 .
Example 4
What is the altitude of a prism ?
Solution:
A prism has two congruent polygons joined together at their corresponding
vertices. The perpendicular distance between the two polygons is the height of the
prism. It is hence obvious that in a right prism the length of the lateral side is the
altitude of the prism.
Example 5
If the volume of a cube is 27. Find the surface area.
Solution:
Volume of a cube = cube of its length
= l3 = 27
Since l 3 = 27
l = 3
Surface area of a cube = 6 l 2
= 6 (3)2
= 6 9
= 54 sq. units
Example 6
If the height of a cuboid is zero it becomes a (a) prism (b) cube (c) rectangle.
Solution:
A cuboid is two congruent rectangles that are separated from each other by a
distance = height. If the height is zero these two will collapse on each other to form
one rectangle.
The other two options i.e. prism and cube are obviously wrong because both
cannot have height = zero.
8.4 Circular Cylinders
Pillars, pipes etc. are examples of circular cylinders encountered daily.
A circular cylinder is two circles with the same radius at a finite distance from each
other with their circumferences joined.
Figure 8.7
The definition of a circular cylinder is a prism with circular bases. The line joining
the centers of the two circles is called the axis.
If the axis is perpendicular to the circles it is a right circular cylinder otherwise it is
an oblique circular cylinder.
The lateral area of a right circular cylinder is the product of the circumference and
the vertical distance between the two circles or the altitude ‘h’.
= Circumference h
= 2r h
= 2 r h square units.
where ’r’ is the radius of the circle.
The total area must include apart from the lateral area, the areas of the two circles.
Area of the circle = r2
Area of two circles = 2r2
Total area = 2rh + 2r2
= 2r (h + r)
= (h + r) square units
where C = circumference.
The volume of a right circular cylinder,
= area of the circle height
= r2 h
= r2h cubic units.
8.5 Pyramids
A pyramid is a polygon with all the vertices joined to a point outside the plane of
the polygon.
If the polygon is regular then the pyramid is called a regular pyramid and is named
by the polygon which forms its base.
If the base is a square the pyramid is called a regular square pyramid, if it is a
pentagon the pyramid is called a regular pentagonal pyramid and so on and so
forth.
The parts of the pyramid are named analogous to the geometric solids mentioned
earlier in the chapter. It is a base, lateral faces, lateral edges and an altitude.
The terminology varies only for the vertex. In all the geometric solids seen so far
all corners are called vertices. In a pyramid however the apex to which all corners
of the polygon are joined is called the vertex of the pyramid.
A regular pyramid has a property called slant height which is the perpendicular
distance between the vertex and any side of the polygon.
The lateral area of a regular pyramid is defined using this parameter.
Lateral area of a regular pyramid = square units
where ’p’ is the perimeter and ’ l’ is the slant height.
Figure 8.8
In pyramid PABC, P is the vertex and PR is the slant height.
Perimeter = l (seg. AB) + l (seg. BC) + l (seg. CA)
= p
8.6 Right circular cone
A right circular cone is a circle with all points on its circumference joined to a
point equidistant from all of them and outside the plane of the circle.
The lateral area of the right circular cone
The total area TA = (LA + BA) square units.
BA = base area = r2
TA = r l + r2
= r ( l + r) square units
8.7 Sphere
The simplest example of a sphere is a ball. One can call here that a circle is a set of
points in a plane that are equidistant from one point in the plane. If this is extended
to the third dimension, we have all points in space equidistant from one particular
point forming a sphere.
The sphere has only one surfaces and its surface area = 4 r2
The volume of a sphere =
Example 1
If the lateral area of a right circular cylinder is 24 and its radius is 2 . What is its
height.
Solution:
Lateral area of a right circular cylinder = 2 r h
Since r = 2 , LA = 4 h.
Given that LA = 24
24 4 h
h = 6 units
Example 2
Find the total area of the right circular cylinder with a radius of 10 units and a
height of 5 units.
Solution:
Total area of a right circular cylinder is 2rh + 2 r2
If r = 10 and h = 5
Total area = 100 + 200 = 300
Example 3
What is the radius of the right circular cylinder with volume 18 cubic units and
height = 2.
Solution:
Volume of right circular cylinder = r2h
where r = radius and h = height
Given that V = 18and h = 2
18 = r2 2
or r2 = 9
r = 3 units
Example 4
If the perimeter and slant height of a regular pyramid are 10 and 3 respectively.
Find its lateral area.
Solution:
For a regular pyramid lateral area, LA = p l
where p = perimeter and l = slant height
Given that p = 10 and l = 3
LA = 10 3 = 15 square units.
Example 5
If the volume of a sphere is 36. Find its radius and surface area.
Solution:
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CHAPTER 9 : CO - ORDINATE GEOMETRY
9.1 Points and Co-ordinates
Figure 9.1
The position of any given point in a plane can be determined only with respect to a
assumed reference point. For practical purposes, the reference point is an
intersection of two perpendicular lines. One of them is horizontal and the other is
vertical. With the help of these two lines and the reference point, the position of
any given point can be determined. As in the above figure location of point P can
be described as, 4 units to the right of the vertical seg OL and 3 units above the
horizontal seg ON. Here the reference point is assumed to be the beginning or the
origin of both, the horizontal as well as the vertical lines. Conventionally having
value zero.
9.2 Co - ordinates and Axes
In order to indicate the location of the point P, if we decide to write the horizontal
units first and the vertical units next, then the position of the point P can be
expressed as P ( 4, 34 ). This pair of numbers is called co-ordinates of the point P ;
and they entirely describe the position of point P in the plane LMNO with respect
to the reference point 0.
Hence, the location of each point in a plane can be expressed with a pair of
numbers called the co-ordinates of that point.
The horizontal line is called the x-axis and the vertical line is called the y-axis. The
reference point is called the origin. The x-axis and y-axis are co-planar. These
points as well as the points in their plane are called a co-ordinate plane ( figure 9.2)
Figure 9.2
X - Co-ordinate: While determining the position of a point in the co-ordinate
plane, the value of the horizontal unit is called the x-co-ordinate of that point. This
value is positive if measured to the right of the origin and negative if measured to
the left of the origin.
Y- Co-ordinate: While determining the position of a point in the co-ordinate
plane, the value of the vertical unit is called the y-co-ordinate of that point. This
value is positive if measured upwards from the origin and is negative if measured
downwards from the origin.
Ordered Pair: In order to name a point in the co-ordinate plane, the x and y co-
ordinates of that point have to be written (first x-co-ordinate then y-co-ordinate) in
brackets / parentheses separated by a coma. This pair of numbers is called the
ordered pair. For e.g. if the x and y-co-ordinates of a point A are 3 and 5
respectively then the ordered pair to describe A is written as A (3, 5). The ordered
pair for the origin is (0, 0).
Example 1
Identify the co-ordinates of the points A, B, C & D in the following figure and
write them in their ordered pairs.
Solution:
A ( 2, 4 )
B ( 5, 7 )
C ( 5, -5 )
D ( -6, 7 )
9.3 Quadrants
The x-axis and the y-axis divide the co-ordinate plane into four regions. These
regions are called as Quadrants. They are shown in the following figure.
i) The upper right quadrant is the 1st quadrant. The values of x co-ordinate and y
co-ordinate of any point in this quadrant is always positive. Its ordered pair is
expressed as ( +, + ).
ii) The upper left quadrant is the 2nd quadrant. Here the value of the x co-ordinate
is negative (measured left from the origin) but the value of the y-co-ordinate is
always positive. The ordered pair is expressed as ( -, + ).
iii) The lower left quadrant is the 3rd quadrant. The values of the x-co-ordinate and
the y-co-ordinate are both negative here. The ordered pair is expressed as (-, -).
iv) The lower right quadrant is the 4th quadrant. In this quadrant the x-co-ordinate
has a positive value and the y-co-ordinate has a negative value. The ordered pair
is expressed as ( +, - ).
9.4 Distances and Distance Formula
With the help of x and y axis we saw how the position of a point in the co-ordinate
plane was determined. We shall now extend this theory to calculate the distances
between any two points in the co-ordinate plane.
Let A (x, y) and B (x2, y2) be two points in the co-ordinate plane as shown below:
Figure 9.4
In order to find the distance between points A and B we go through following steps
and use the distance formula.
Step 1 Draw a line parallel to x-axis through the point A and draw a line parallel to
y-axis through point B such that they intersect at point C.
Step 2 We now have a right triangle with seg AB as the hypotenuse. Therefore, by
Pythagorean theorem, (AB)2 = (AC)2 + (BC)2
AB =
Step 3 The co-ordinates of point C can be determined as (x2, y1).
Step 4 The distance between A and C is, AC = (x2, x1)
(AC)2 = (x2, x1)2 and The distance between B & C is
BC = (y2, y1)
(BC)2 = (y2, y1)2
Step 5 Substituting the values of (AC)2 and (BC)2 in eg (1) we get
This is the distance formula.
Theorem 1: If the co-ordinates of two points are (x1, y1) and (x2, y2), then the
distance d between them is given by the formula:
Example 1
Using the distance formula determine the distance between the points with co-
ordinates ( 4, -2 ) and (6, 3 ).
Solution:
Let (4, -2) be (x1, y1) and
(6, 3) be (x2, y2). Then according to the distance formula,
Example 2
A triangle has vertices A (2,2), B (5,2) and C (5,6). Show that the triangle is a right
angled triangle.
Solution:
By the distance formula
d(AC) = 5, d (AB) = 3, and d (BC) = 4
According to Pythagorean if the sum of the square of two sides are equal to the
square of the third side then the triangle is a right angled triangle.
Therefore, the triangle is right angled triangle.
9.5 Mid point Formula
Theorem: If the co-ordinates of the end points of a segment are (x1, y1) and (x2,
y2), then the co-ordinates of the mid point of this segment is given by the formula:
Example 1
Find the mid point of a segment if A is (6,8) and B is (-2, 4).
Solution:
Let (6, 8) be (x1, y1) and
(-2, 4) be (x2, y2).
Therefore their mid point M is given as :
Example 2
If the mid point of seg. AB is (-3, 8) and A is (12, -1), find co-ordinates of B.
Solution:
Let the co-ordinates of B be (x, y). According to the mid-point formula
Therefore the co-ordinates of B are (-18, 17).
9.6 Slope of a Line
As it is known any two given points contain a line. There can be various
lines on a co-ordinate plane. Some of them parallel to the x-axis and some
inclined to it. To answer the question - which way they incline and how
much, is the concept of the slope of a line.
The slope of a line can be said to be the measurement of steepness of the
line and its direction.
The line is said to have a positive slope if it rises from left to right as shown
in the following figure 9.5 (a).
9.5(a)
9.5(b)
The line is said to have a negative slope if it rises from right to left (figure 9.5(b).
The slope of a line is zero if it is parallel to the x-axis and it is undefined if it is
parallel to the y-axis (see figure 9.6 (a) and (b)).
Slope : If a line passes through points A & B with the co-ordinates ( x1, y1) & (x2,
y2) then the slope of the line AB can be represented as
If a line is parallel to the x-axis then its y-co-ordinate is the same for all the points
on the line.
Therefore the slope of a line parallel to the x-axis is 0.
Slope of Parallel Lines:
Theorem: If two non vertical lines are parallel, then they have the same slope.
OR
If the two lines have the same slope then the lines are parallel and non vertical.
Figure 9.7
Slope of Perpendicular lines:
Theorem: If two non-vertical lines are perpendicular to each other, then their
slopes are opposite reciprocals of each other.
Figure 9.8
If the product of slopes of two lines is -1, then the lines are non-vertical and
perpendicular to each other.
Example 1
If a line l has a slope of 5/4 then (a) find the slope of a line parallel to l (b) find
slope of a line perpendicular to l .
Solution:
a) The slope of line parallel to l is 5/4.
b) Let the slope of line perpendicular to be x.
9.7 Equation of a line
Observe the following co-ordinate plane for the values of the marked points.
The observations are:
a) The co-ordinates of points B & C are B (-2, 1) & C (2, 1)
b) The co-ordinates of point A & C are A (2, -2) & C (2, 1 ).
Note that the x- co-ordinate value on the line l is the same for all the points on that
line. This means that it satisfies the equation x = 2 for all its points.
Similarly the equation y = 1 holds true for all the points on the line m in both the
cases above. We have considered lines that were parallel to the x-axis and the y-
axis and hence their equations appear in one variable. If a slanted line intersecting
both x and y-axis is considered then the equation would involve two variables and
the equation can be written in the form of Ax + By = C, where A, B & C are real
numbers.
In general, an equation of the type Ax + By = C can be graphed on a co-ordinate
plane by substituting different values for x and y. If the graph is a straight line then
the equation is said to be linear.
If any other point lies on this line, then its co-ordinate will make the equation a true
statement. This form of an equation is called the standard form for the equation of
the line.
X- Intercept and Y- Intercept: The x-intercept of a graph is the point where the
graph intersects the x axis. The y co-ordinate of this point is always zero.
**********