#### Geometry concept

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### 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.

An infinite number of lines can be drawn through any given point.

One and only one line can be drawn through two distinct points.

When two lines intersect they do so at only one point.

**Collinear And Coplanar**

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.

figure1.4

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 drawn namely AB, BC & AC.

**Example 2**

Figure 1.7

From figure 1.7 answer the following :

Name lines parallel to line AB

Are line AO and point R coplanar ? Why ?

Are points A, S, B and R coplanar ? Why ?

Name three planes passing through at A.

Solution :

Line CD, line SR and line PQ.

Yes. Any line and a point outside it are coplanar.

Yes. Two parallel line are always coplanar.

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.

**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 Ð and named, using either both the rays or just the vertex (figure 1.11).

Figure 1.11

Figure 1.11 represents Ð XYZ or Ð Y.

**Interior and Exterior of an angle :**The interior of Ð PQR is the shaded region in figure 1.12. S is a point in the interior of ÐQ 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 ÐPQR.Figure 1.12

In figure 1.13 the shaded region shows the exterior of Ð 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 0^{0}to 180^{0}and is represented as m Ð. 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 Ð AOB and m Ð AOB = 180

^{0}Figure 1.15

All rays starting from O and going above line AB form angles with such that their measure is between 0

^{0 }and 180^{0}. eg. Ð COB , Ð DOB and Ð 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 Ð COB + m Ð DOC = m Ð 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 ) Ð AOB and Ð COB are equal in measure then is called the bisector of Ð 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 90

^{0}. It is represented by the symbol . Since its measure is fixed, it follows that all right angles are equal. Ð AOB (figure 1.17 ) is a right angle.Figure 1.17

**Acute angle**: Any angle whose measure is between 0 and 90^{0}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 0^{0}and 180^{0}is called an obtuse angle (figure 1.19).Figure 1.19

90

^{0}< b < 180^{0 }\ b is an obtuse angle.**1.4 Some special angles****Complementary angles :**If the sum of two angles equals 90^{0}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 90

^{0}if the measure of one angle is known that of the complement can be found easily. In figure 1.20 b Ð a and Ð b are complementary. Also it is known that m Ð a = 30^{0}.m Ð a + m Ð b = 90

^{0}30

^{0}+ m Ð b = 90^{0}m Ð b = 90

^{0}– 30^{0}m Ð b = 60

^{0}**Theorem :**If two angles are complementary to a third angle, then they are equal to each other.**Proof :**Ð a and Ð b are both complementary to Ð c.\ m Ð a + m Ð c = 90

^{0}and alsom Ð b + m Ð c = 90

^{0}\ m Ð a + m Ð c = m Ð b + m Ð c OR

m Ð a = m Ð b

**Theorem :**If two angles are complementary to equal angles they are equal to each other . If Ð a and Ð b are complementary to Ð c and Ð d respectively where m Ð c = m Ð d.**Proof :**m Ð a + m Ð c = m Ð b + m Ð d = 90^{0}m Ð a + m Ð c = m Ð b + m Ð d

Since m Ð c = m Ð d

m Ð a + m Ð c = m Ð b + m Ð c

or m Ð a = m Ð b.

**Supplementary angles :**If the measures of two angles sum up to 180^{0}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 Ð 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 Ð AOC and Ð BOD. It is observed that and are opposite rays and so is and . In such a case Ð AOC and Ð BOD are called vertical angles ( figure 1.23 ).**Theorem :**Vertical angles are always equal in measure.**Proof :**To prove m Ð AOC = m Ð BODm Ð AOC + m Ð COB = 180

^{0}( supplementary angles )m Ð BOD + m Ð COB = 180

^{0}( supplementary angles )i.e. m Ð AOC + m Ð COB = m Ð BOD + m Ð COB

or m Ð AOC = m Ð BOD.

Figure 1.23

**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

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.**l****Corresponding angles :**Angles that appear in the same relative position in each group are called corresponding angles, i.e. Ð 1 and Ð 5 are called corresponding angles. Similarly Ð 2 & Ð 6, Ð 4 & Ð 8 and Ð 3 & Ð 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. Ð 3, Ð 4, Ð 5 and Ð 6 . Exterior angles lie on opposite sides of lines a and b, i.e. Ð 1, Ð2, Ð 7 and Ð 8 .**Alternate and Consecutive Interior angles :**Interior angles on opposite sides of the transversal are called alternate interior angles and Ð 4, Ð 6 are alternate interior angles and so also Ð 3, and Ð 5 .Interior angles on same side of the transversal are called consecutive interior angles. Ð 4, and Ð 5 are consecutive interior angles and so also Ð 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. Ð 1 and Ð 7 and also Ð 2 and Ð 8 .Exterior angles on the same side of the transversal are called consecutive exterior angles, i.e. Ð 1 & Ð 8 as also Ð 2 and Ð 7

**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 Ð 1 = m Ð 5

m Ð 2 = m Ð 6

m Ð 3 = m Ð 7

m Ð 4 = m Ð 8 .

**Alternate interior angles :**If a transversal cuts two parallel lines the alternate interior angles are equal in measure. In figure 1.26 m Ð 4 = m Ð 6 and m Ð 3 = m Ð 5. This can be proved as follows :m Ð 3 = m Ð 4 are supplementary and so also

m Ð 6 = m Ð 7 are supplementary.

Since the sums of their measures are 180

^{0}in both cases.m Ð 3 + m Ð 4 = m Ð 6 + m Ð 7.

However m Ð 3 = m Ð 7 as they are corresponding angles formed by a transversal across parallel lines.

Therefore, m Ð 3 + m Ð 4 = m Ð 6 + m Ð 3 i.e. m Ð 4 = m Ð 6.

Similarly it can be shown that m Ð 3 = m Ð 5 .

**Alternate exterior angles :**If a transversal cuts two parallel lines the alternate exterior angles are equal in measure. In figure 1.26 m Ð 1= m Ð 7 and m Ð 2 = m Ð 8. This can be easily shown as follows :m Ð 1 = m Ð 3 (vertical angles)

m Ð 3 = m Ð 7 (corresponding angles on parallel lines)

\ mÐ 1 = m Ð 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 Ð 4 and m Ð 5 are supplementary and also m Ð 3 and m Ð 6 are supplementary. This is proven as :m Ð 1 and m Ð 4 are supplementary

but m Ð 1 = m Ð 5 (corresponding angles on parallel lines )

\ m Ð 4 and m Ð 5 are supplementary.

Since consecutive exterior angles are supplementary. m Ð 1 and m Ð 8 so also m Ð 2 and m Ð 7 should be supplementary.

**Proof :**m Ð 1 and m Ð 4 are supplementary butm Ð 4 = m Ð 8 (corresponding angles on parallel lines )

\ m Ð 1 and m Ð 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 show 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

**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 D and is named by its vertices as shown in figure 2.1.

Figure 2.1

D ABC has three sides AB, BC and CA. It has three angles

Ð ABC, Ð BCA and ÐCAB**2.2 Sum of the angles of a triangle**It can be proven easily that the sum of the three angles of a D is 180

^{0}Figure 2.2

D 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 Ð PAB and m ÐABC are equal as they are alternate interior angles. Similarly m Ð QAC = m Ð ACB.

Now Ð PAQ = m Ð PAB + m Ð BAC + m Ð CAQ

i.e. 180

^{0}= m Ð ABC + m Ð BAC + m Ð ACBm Ð PAQ = 180

^{0}because it is a straight line. Thus the sum, of the measures of the three angles, of any triangle, is 180^{0}.**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.

**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

**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

**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****Acute triangle :**A triangle in which all the angles are acute, ( i.e. < 90^{0}) 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 D 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 180

^{0}, it can be said that each angle of an equiangular triangle is 60^{0}.**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 180

^{0}it can be said that the other two angles of an obtuse triangle are acute.**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 Ð KJL is 90

^{0}it can be said that Ð JKJL and Ð 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 .

**Altitudes of an acute triangle :**Figure 2.10

For an acute triangle figure 2.10 all the altitudes are present in the triangle.

**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

**Altitudes for an obtuse triangle :**Figure 2.12

D 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.

**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 D PQR @ D 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 criterias 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

\ D ABC @ D 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 Ð ABC = m Ð PQR

\ D ABC @ D 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 Ð B + m Ð R m Ð L = m Ð P and seg. BC = seg. RP

\D ABC @ D 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 Ð A = m Ð P m Ð B = m Ð Q and AC = PR

\ D ABC @ D 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

\ D ABC @ D PQR by HS postulate.

**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 D ABC where AB = AC ( figure 2.23 ).

Figure 2.23

**Proof : **To prove m Ð B = m Ð C drop a median from A to BC at point P. Since AP is the median, BP = CP.

\ In D ABP and D 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.

D ABP @ D ACP

\ m Ð B = m Ð C as they are corresponding angles of congruent triangles.

The converse of this theorem is also true and can be proven quite easily.

Consider D ABC where m Ð B = m Ð 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 Ð BAP = m Ð CAP

m Ð ABP = m Ð ACP ( given )

seg. AP @ seg. AP (same side )

\ D ABP @ D 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.

**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 vertice

**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 180^{0}, the sum all the internal angles of this polygon is 6 ´ 180^{0} = 1080^{0}.

This can be generalized as :

For any n sided polygon the sum of its internal angles is

( n – 2 ) ´ 180^{0}.

**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 ) ´ 180^{0}

= 3 ´ 180^{0}

= 54^{0}

\ each interior angle of the pentagon measures

540^{0} / 5 = 108^{0}

The interior and exterior angles form linear pairs and hence are supplementary.

\ Each exterior angle measures 180^{0} – 108^{0} = 72^{0}

\ Sum of five exterior angles = 5 ´ 72 = 360^{0}

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 ) 180^{0}.

\ 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 360

**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 Ð MNO.

Consider D LOP & D 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 paralles lines is always the same.

\ By HS postulate D LOP @ D MNQ.

\ Ð LOP @ Ð 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 D LON & D MNO.

seg. LO = seg. MN by definition of isosceles trapezoid.

Ð LON @ Ð MNO base angles are equal. seg. ON + seg. NO same side.

\ D LON @ D MNO by SAS postulate

\ seg. LN @ 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 D 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 Ð MLO = 108 and seg. MO = 4.9.

Find (a) m Ð LMN , (b) m Ð MNO and (c) seg. LN .

Solution:

a) m Ð LMN = 180^{0} as the base angles of an isosceles trapezoid are equal.

b) m Ð MNO = 72^{0} The consecutive interior angles formed by a transversal seg. MN on two parallel lines seg. LM & seg. NO are supplementary. Therefore ÐLMN and ÐMNO 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 çç seg.DC and seg.AD is parallel to seg.BC i.e. seg.AD çç seg.BC. Therefore ABCD is a parallelogram. It is represented as parallelogram ABCD. Since both sides the 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 D ACB and D CAD.

Ð CAB @ Ð ACD ( alternate angles )

Ð ACB @ Ð CAD ( alternate angles )

and seg.AC @ seg.CA ( same side )

\ D ACB @ D 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 D ACB @ D CAD

Ð ABC @ Ð CDA

By drawing a diagonal from D to B it can be shown that Ð DAB @ Ð 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 D SOR and D QOP, Ð OSR @ Ð OQP and Ð ORS @ Ð OPQ (alternate angles ).

SR @ PQ opposite sides of a parallelogram.

\ D SOR @ D 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 are 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 D ADC & D CBA

AD @ CB and DC @ BA ( given ) also AC @ CA ( same side )

\ D ADC @ D CBA ( SSS ).

\ Ð ACB @ Ð CAD and Ð ACD @ Ð CAB ( corresponding angles of congruent triangles are congruent ).

Ð ACB @ Ð CAD Þ AD çç 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 Ð ACD @ Ð CAB AB çç 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 D POS and D ROQ.

seg.PO @ seg.RO and seg.OS @ seg.OQ ( given )

Ð POS @ Ð ROQ ( vertical angles )

D POS @ D ROQ

Ð OPS @ Ð 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 çç QR. Similarly by showing that Ð OSR @ Ð OQP, it can be shown that PQ çç 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.LM@ seg.NO . To prove that LMNO is a parallelogram.

Figure 3.19

Since seg.LM çç seg.ON Ð LMO @ Ð NOM ( alternate angles )

In D LMO and D NOM

seg.LM @ seg.NO

Ð LMO @ Ð NOM

seg.MO @ seg.OM (Same side )

LMO @ D NOM

\ Ð LOM @ Ð 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 çç 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 90^{0} 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 D ACD and D BDC . Both are right triangles.

seg.AD @ seg.BC by definition

seg.DC @ seg.CD same side

Ð ADC @ Ð BCD – both are right angles.

\D ACD @ D 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 D LNO and D MON.

seg.LN @ seg.MO given seg.LO @ seg.MN opposite sides of a parallelogram and seg.ON @ seg.NO same side

\ D LNO @ D MON ( SSS )

Ð LON @ Ð MNO ( corresponding angles of congruent triangles ). Since they are interior angles of parallel lines they are supplementary.

\ Ð LON and Ð MNO are both right angles.

\ Ð 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 D BOA and D BOC.

seg.BA @ seg.BC definition of rhombus …………………….. (1)

Ð ABD @ Ð CDB alternate angles. Ð DBC @ Ð CDB as seg.CD and seg.CB are congruent. Therefore D CBD is isosceles .

\ Ð ABD @ Ð DBC which is the same as Ð ABO @ Ð CBO …………….. (2)

seg.BO @ seg.BO same side …………………….. (3

\ D BOA @ D BOC

\ Ð BOA @ Ð BOC.

Corresponding angles of congruent triangles are congruent.

But Ð BOA and Ð 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 Ð A = 60^{0}, find m Ð B, m Ð L and m Ð D.

Solution:

m ÐB = 120^{0} – The consecutive angles are supplementary.

m ÐL = 60^{0} – The opposite angles are equal.

m ÐD = 120^{0 }– The opposite angles are equal.

**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 cm^{2} 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 = a^{2 }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 D ALD and D BMC. Both are right triangles.

*l* (AD) = *l* (BC) opposite sides of a parallelogram.

*l* (AL) = *l* (BM) altitudes of a parallelogram.

\ D ALD @ D 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 D PQR or

Area D PQR = Area of parallelogram POQR.

Area of parallelogram POQR = bh where h is the altitude on the base with length h.

\ Area D 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 + b_{1} + a_{1} + b

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 + a_{1} ) 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 p r where r is the radius, p = and has been discovered by ancient Greeks, by actually dividing the circumference of a circle by its diameter .

Area of a circle = p r^{2} 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 cm^{2}

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 cm^{2}

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 DABC & D LMN

Solution:

Perimeter D ABC = 3 + 4 + 5 = 12 cm

Area of D ABC = ´ 4 ´ 3 = 6 cm^{2}

Perimeter D LMN = 36 + 36 + 52 = 124 ft.

Area of D 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 cm^{2}

Example 9

Find the circumference of a circle with area 25 p sq. ft.

Solution:

Area of the circle = p r^{2} = 25 p

\ r = 5 ft.

\ circumference = 2 pr

= 10 p

Example 10

Find the area of a circle with circumference 30p cm.

Solution:

Circumference = 2 p r = 30 p

r = 15 cm

area = p r^{2}

= p ´ 15^{2}

= 225 p

**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 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 180^{0}x + 2x = 180^{0} 3x = 180^{0} x = 60^{0} \ 2x = 120^{0}The two angles are 60^{0} and 120^{0}

**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 substraction 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 D PTS and D QTS

Areas of triangles with same height are in the ratio of their bases.

Similarly

But A ( DQTS ) = A ( D 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 D ABC ( figure 5.5) the bisector of Ð 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 çç seg.DA

Ð BAD @ Ð AEC , corresponding angles

Ð DAC @ Ð ACE , alternate angles

But Ð BAD = Ð DAC , given

\ Ð AEC @ Ð ACE

Hence D AEC is an isosceles triangle.

\ seg.AC @ seg.AE

In D BCE AD çç CE

Thus the bisector divides the opposite side in the ratio of the remaining two sides

Example 1

In the diagram seg.PQ çç 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 çç seg.DC. PQ çç 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 D ABC, seg.BP is the bisecter of Ð 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 another triangles, then two triangles are similar.

Example 1

D ABC and D 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 D ABC ~ D DEF ~ D 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 D ABC ~ D DEF ~ D 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 . D LMN and D PQR.

Figure 5.6

This ratio is called the scale factor.

Perimeter of D LMN = 8 + 7 + 10 = 25

Perimeter of D 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 ~ D PQR

To prove that

Draw perpendicular from A and P to meet seg.BC and seg.QR at D and S respectively.

Since D ABC ~ D PQR

also Ð B @ Ð Q

In D ABD and D PQS

also Ð B @ Ð Q and Ð ADB @ Ð PSQ

\ D ABD ~ D 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 D PQR whose area is four times area of D 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 D ABC ~ D PQR such that l (seg.AB) : l (seg.PQ) that is 8 : 6. If area of D ABC is 48 sq.cm what is the area of the smaller triangle. Solution: Example 5 In a trapezium ABCD, side AB çç CD. The diagonals AC and BD cut each other at M. Prove that Solution: To Prove that Consider D AMB and D CMD. Ð AMB @ Ð CMD, vertical angles Ð BAM @ Ð DCM alternate angles \ By AA test D AMB ~ D CMD \ |

**CHAPTER 6 : THEOREM OF PYTHAGORAS AND THE RIGHT TRIANGLE**

**6.1 The Right Triangle**

Figure 6.1

D ABC is a right triangle, hence m Ð ABC = 90^{0}. Therefore m Ð A and m Ð C are complementary ( figure 6.1). Now seg.BD is a perpendicular onto seg.AC (figure 6.2).

Figure 6.2

Seg.BD divides D ABC into two right triangles D BDC and D ADB ( figure 6.2). It can be easily proven that these two triangles are similar to the parent D ABC and therefore similar to each other.

Proof :

Consider D ABC and D BDC

Ð ABC @ Ð BDC right angles and

Ð BCA @ Ð DCB same angle

\ by AA test D ABC ~ D BDC ® (1)

Similarly consider D ABC and D ADB.

Ð ABC @ Ð ADB right angle

Ð CAB @ Ð DAB same angle

\ by AA test D ABC ~ D ADB ® (2)

from (1) and (2)

D ABC ~ D ADB ~ D BDC.

Since D ABC ~ D BDC

and D ABC ~ D 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 DADB and DBDC 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 D 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 D BDC and D 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.

\ x^{2} = 12 ´ 3

x^{2} = 36

x = 6.

Example 3

Find y.

**Solution :**

**6.2 The Theorem of Pythagoras**

Figure 6.3

D ABC is a right triangle.

l (AB) = c

l (BC) = a

l (CA) = b

CD is perpendicular to AB such that

D ABC ~ D CBD

or l (BC)^{2} = l (AB) ´ l (CD)

a^{2} = c ´ x = cx® (1)

D ABC ~ D ACD

or l (AC)^{2} = l (AB) ´ l (AD)

b^{2} = c ´ y = cy ® (2)

Therefore, from (1) and (2)

a^{2} + b^{2} = cx + cy

= c ( x + y )

= c ´ c

= c^{2}

a^{2 }+ b^{2} = c^{2}

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 D ABC l (AC)^{2} = l (AB)^{2} + l (BC)^{2}

To prove that D ABC is a right triangle or that

m Ð ABC = 90^{0}.

Construct a D PQR such that

l (QR) = l (BC) , l (PQ) = l (AB) & Ð PQR = 90^{0}.

In D 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 D ABC and D PQR

l (AB) = l (PQ) construction

l (BC) = l (QR)construction

l (AC) = l (PR) from (1)

\ D ABC @ D PQR by S S S.

\ M Ð B = M Ð Q

Since m Ð Q = 90^{0} Ð B is a right angle and D 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 Ð ABC is 90^{0}.

D 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

D 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**

D ABC is a right triangle. m Ð ACB = 90^{0} , seg.AQ bisects seg.BC at Q.

Prove that 4 l (AQ)^{2} = 4 l ( AC)^{2} + BC^{2}

Solution :

D 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

\ 4 l (AQ)^{2} = 4 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 c^{2} > a^{2} + b^{2} D ABC is an obtuse and if c^{2} < a^{2} + b^{2} D ABC is an acute triangle.

**6.3 Special Right Triangles**

**The 30 ^{0} – 60^{0} – 90^{0} triangle :** If the angles of a triangle are 30

**, 60**

^{0}**and 90**

^{0}**then the side opposite to 30**

^{0}**is half the hypotenuse and the side opposite to 60**

^{0}**is times the hypotenuse.**

^{0}Figure 6.5

In D ABC Ð A = 90^{0} Ð B = 60^{0} and Ð C = 30^{0}

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 D ABC and D ACD

AB @ AD construction

Ð CAB @ Ð CAD both are right angles

AC @ AC common side

\ D BAC @ D DAC (SAS)

\ Ð ACB @ Ð ACD corresponding angles of congruent triangles are equal.

but m Ð ACB = 30^{0}

\ m Ð DCB = 60^{0}

Þ D 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)^{2}from (3)

l (seg.AC)^{2} = l (seg.BC)^{2} – l (seg.BC)^{2}

= l (BC)^{2}

**The 45 ^{0} – 45^{0} – 90^{0} triangle :** If the angles of a triangle are 45

^{0}– 45

^{0}– 90

^{0}then the perpendicular sides are times the hypotenuse.

In D ABC Ð A = 45^{0} , Ð B = 90^{0} and Ð C = 45^{0}.

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) . D 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 Ð LON = 30^{0} 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 Ð LON = m Ð MNP = 30^{0}

D MNP is a 30^{0} – 60^{0} – 90^{0} triangle

\ l (seg.MP) = l (seg.MN)

l (seg.MN) = l (seg.LO) = 12 cm

\ l seg.MP = ´ 12 cm

= 6 cm.

Example 2

D 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 D 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 D 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) – 2 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 8.1 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 8.1 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 7.2).

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 7.2 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 Ð 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 Ð BAC is the inscribed angle and arc BXC is the intercepted arc. To prove that m Ð BAC = 1/2 m (arc BXC). There arise three cases as shown in figure 8.6 (a), 8.6 (b) and 8.6 (c).

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.

D OAC is an isosceles triangle as seg.OA = seg.OC.

Assume m Ð OAC = m Ð OCA = P

m Ð COA = 180 – 2 P as sum of all the angles of a triangle is 180^{0}.

Ð COA & Ð COB form a linear pair. Therefore they are supplementary.

m Ð COB = 180^{0} – m Ð COA

= 180^{0} – ( 180^{0} – 2P)

= 2P

= 2 m Ð BAC

But m Ð BAC = m (arc BXC)

\ m Ð 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 Ð CAD and let arc Ð BND be that which is intercepted by Ð DAB.

From case 1 m Ð CAD = m ( arc CMD)

and m Ð DAB = m (arc BND)

m Ð CAD + m Ð DAB = m (arc CMD) + m (arc BND)

m Ð 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 Ð CAD and let arc BND be the one intercepted by Ð DAB.

From case 1 m Ð CAD = m (arc CMD)

m Ð DAB = m (arc BND)

m Ð CAD – m Ð DAB = m (arc CMD) – m (arc BND)

m Ð 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 Ð CAB and Ð CDB intercept the same arc CXB.

Prove that m Ð CAB = Ð CDB.

From the previous theorem it is known that

m Ð CAB = m (arc CXB) and also

m Ð CDB = m (arc CXB)

\ m Ð CAB = m Ð 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 90^{0}.

Figure 8.8

The inscribed angle Ð ACB intercepts a semicircle arc AXB (figure 8.8). We have to prove that m Ð ACB = 90^{0}.

m Ð ACB = m (arc AXB)

= (180^{0})

= 90^{0}

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) Ð AOB

b) 80^{0}. 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 – Ð PRQ

intercepted arc – arc PQ

b) 60^{0}. The measure of an intercepted arc is twice the measure of its inscribed angle.

Example 3

Ð PAQ and Ð PBQ intercept the same arc PQ what is the m Ð PBQ and m (arc PQ) ?

Solution:

m Ð PBQ = 40^{0} If two inscribed angles intercept the same arc their measures are equal m (arc PQ) = 80^{0} 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 D PDB and D PAB

seg.BD @ seg.BA ( B is the midpoint of AD)

ÐPBD @ ÐPBA ( PB is perpendicular to l ) and

seg.PB = seg.PB (same segment)

\ D PBD @ D 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 Ð1 and Ð 2.

Figure 7.10

m Ð1 = (m seg.AD + m seg.BC) and

m Ð2 = (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 Ð ABX and Ð ABY are half the measures of the arcs they intercept.

m Ð 1 = m (arc ACB)

m Ð 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 Ð ABC intercepts a semicircle the chord AB passes through the center. Therefore m Ð ABC = 90^{0} (a tangent is always perpendicular to the diameter that intersects it at the point of tangency).

m (arc BXA) = 180^{0} (arc BXA is a semi circle)

\ m (arc BXA) = ´ 180^{0} = 90^{0}

\ m Ð ABC = m (arc BXA)

**Case 2 :** Assume that ÐABC intercepts a minor arc. Therefore as seen in figure 8.13 the center O lies in the exterior of Ð ABC.

Figure 7.13

m Ð ABC = 90^{0} – m Ð ABO

m Ð ABO = 90^{0} – m Ð ABC ———– (1)

But m Ð ABO = m Ð OAB

(as OAB is an isosceles triangle )

\ m Ð OAB = 90^{0} – m Ð ABC ———– (2)

(1) + (2)

\ m Ð ABO + m Ð OAB = 180 – 2 m ÐABC

Since m ÐABO + m Ð OAB = 180 – m Ð BOA

180 – m Ð BOA = 180 – 2 m Ð ABC

i.e. m Ð BOA = 2 m Ð ABC

m Ð ABC = m Ð BOA

m Ð ABC = m ( arc AXB )

**Case 3 :**

Figure 7.14

If Ð ABC intercepts a major arc, the center of the circle O will lie in the interior as ÐABC . See figure 7.14.

Now ÐADB intercepts a minor arc AYB.

\ m Ð AOB = m (arc AYB)

\ 180^{0} – m ÐADB = { 360^{0} – m (arc AXB) }

\ 180^{0} – m ÐADB = 180^{0} – m (arc AXB)

\ m Ð 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 .

Ð 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 Ð AXD = 115^{0} and m (arc CB) = 45^{0} . Find m arc APD.

Solution:

m arc APD = 185^{0}

m Ð AXD = { m (arc APD) + m (arc CB) }

m ( arc APD) = 2 m Ð AXD – m (arc CB)

= 2 ´ 115^{0} – 45^{0}

= 185^{0}

Example 2

l is a tangent to the circle at B. Seg. AB is a chord such that m Ð ABC = 50^{0}. Find the m (arc AB).

Solution:

m (arc AB) = 100^{0}

m Ð ABC = m (arc AB)

50 = m (arc AB)

m (arc AB) = 100^{0}

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 Ð PAQ = 25^{0} and m Ð ROS = 80^{0} find m (arc PQ).

Solution:

m ( arc PQ) = 30^{0}

m Ð PAQ = { m (arc RS) – m (arc PQ)

2 m Ð PAQ = m (arc RS) – m (arc PQ)

\ m (arc PQ) = m (arc RS) – 2 m Ð PAQ

= 80^{0} – 50^{0}

= 30^{0}

**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 Ð AOB = m Ð COD.

Figure 7.16

To prove this, join A, B, C & D withO.

Consider D AOB and D COD

seg.AO @ seg.CO and seg.OB @ seg.OD (radii of a circle are always congruent).

seg.AB @ seg.CD (given)

\ D AOB @ D COD ( S S S )

\ Ð AOB @ Ð 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 D LOM and D POQ

seg. OL @ seg. OP and seg. MO @ seg QO ( as all are radii ).

Ð LOM @ Ð POQ (given )

\ D LOM @ D 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 D OXP and D OXY

Both are right triangles.

hypotenuse seg. OX @ hypotenuse seg. OY

( both are radii of the circle )

seg. OP @ seg OP (same side)

\ D OXP @ D 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 that l (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 D POX and D ROY

seg. PX @ seg.RY

seg.OP @ seg.OR. Both being radii onto the circle.

\ D POX @ D 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 D OIP and D 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).

\ D OIP @ D 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 Ð COD = 95^{0}. Find m arc AB.

**Solution:**

95^{0}

m arc AB = m Ð AOB

Since D AOB @ D COD by SSS m Ð AOB = m Ð 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 = 80^{0}, find l (PQ) and m arc PR.

**Solution:**

l (PQ) = 3

m (arc PR) = 40^{0}

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)

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 DAPC and DBDP.

m Ð APC = m Ð BPD (vertical angles).

m Ð CAP = m Ð BDP (angles inscribed in the same arc).

\ DAPC ~ D 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 D PAC and D PBD

m Ð APC = m Ð BPD (same angle).

m Ð CAP = m Ð PDB (exterior angle property of a cyclic quadrilateral).

\ D PAC ~ D 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 D PTA and D PTB.

m Ð TPA = Ð TPB ( same angle)

According to Tangent Secant theorem,

m Ð ATP = m (arc AT)

= m Ð PTB ( inscribed angle )

\ D PTA ~ D 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.

Figure 7.23

To prove that l (seg.QR) = l (seg.QS) join P to Q and R to S.

m Ð PRQ = m Ð PSQ = 90^{0}.

The radius and the tangent form a right angle at the point of tangency,

\ D PRQ and D PSQ are right triangles such that

seg. PR @ seg PS (radii of the same circle).

seg. PQ @ seg. PQ (same side).

\ D PRQ @ D 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 and l (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 360^{0} corresponds to the full circumference. i.e. 2 p r then for a central angle of x^{0} (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. 360^{0} corresponds to area of the circle p r^{2}. 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 75^{0}. Find l(seg.AB). Also find the area of the sector AOB having a central angle of 75^{0}

Solution:

l (seg AB) = 2.6