Ganit Learning Guides Intermediate Geometry lines in a plane can have more than one point in common....

Ganit Learning Guides

Intermediate Geometry Lines, Angles, Triangles, Loci, Concurrency, Quadrilaterals, Parallelograms, Areas of Plane Figures, Surface Area and Volume of Solid Figures, Graphs and Geometric Construction

Author: Raghu M.D.

Intermediate-Geometry 1 of 96 ©2014 www.learningforkowledge.com/glg

GEOMETRY

LINES, ANGLES AND TRIANGLES

Geometry is a branch of mathematics which covers, points, lines, angles, areas and solids.

The word geometry is derived from the Greek words ‘geo’ meaning ‘earth’ and ‘matrein’

meaning ‘measurement’. Some concepts of geometry are as follows.

Postulate or Axiom: Postulates are self evident specific assumptions made to prove

geometrical results or relationships.

Theorem: Theorem is the result derived on the basis of postulates and common logic.

Point: A point is the smallest possible dot in a plane. A point has no length, breadth or

thickness. Points are the basic building blocks of lines and curves.

Lines: Lines are made up of a number of points positioned side by side. A line is a

geometrical figure which has length but no breadth.

A line drawn between 2 points is known as a line segment. A line is considered as a

straight line if the distance between any two points, is the least. Only one line can pass

through two points but an infinite number of lines can pass through one point. All points

on a line are called collinear points. Two distant lines on a plane cannot have more than

one point in common. This point is called as the point of intersection and the lines

intersecting lines. Some pairs of lines don’t intersect at all and are called parallel lines.

Also from a point, not in the same line, there is only one line which passes through the

point and parallel to the given line. A line with only one point defined is called a ray.

A system of rays passing through points BA and

A B

BA and : Two points

AB: Line segment.


Intersecting and parallel lines.

E A F

B O C

D

A: A point in the plane

BC: A line in the same plane as point A

AD and BC: Intersecting lines

O: Points of intersection

BC and EF: A pair of parallel lines

ANGLES

Angle: Angle is defined as the union of two lines (two rays) having the same end point.

Angles are classified according to the degree of separation of the two intersecting rays.

End point where the two rays meet is known as vertex. There can be more than 2 rays

emerging from a vertex resulting in the formation of more than one angle. Properties of

angles are shown in figures 8.3, 8.4, 8.5

Angle ABC formed by ray AB and BC with point B as the vertex

A

B C

Fig 8.4: Three rays AO, BO and CO emerge from the vertex O. Angles AOB and BOC

are the two adjacent angles. Adjacent angles add up to equal the angle formed by the two

extreme rays.

A O B + B O C = A O C


A

B

O C

Property: Adjacent angles on a line are supplementary

Fig 8.5: A ray is drawn from a point A such that it intersects the line (ray) BC at point O

The adjacent angles A O B+A O C = B O C = 180º

A

B O C

Property: If two lines (rays) intersect, vertically opposite angles are equal.

AB and CD are line intersecting at point O. Angles A O C and B O D, B O C and A O D are

pairs of vertically opposite angles.

A O C = B O D, B O C = A O D

A D

O

C B

Proof:

Angles A O D and B O D are adjacent angles on line AB

Hence A O D + B O D = 180º

Angles B O D and B O C are adjacent angles on line CD

Hence B O C+B O D =180º


(A O D+B O D) – (B O D + B O C) = 180º-180º

A O D+B O D - B O D - B O C = 0

A O D - B O C = 0

A O D+B O C like wise A O C - B O D

TYPES OF TRIANGLES

Acute angle: An angle whose measure is less than 90º is called an acute angle

Right angle: An angle whose measure is exactly 90º

Straight angle: If both the rays forming an angle are exactly opposite. A straight angle

has a measure of 180º

Obtuse angle: an angle whose measure is more than 90º but less than 180º

Reflex angle: Reflex angle is also called the exterior angle of an acute, right or obtuse

angle. This is an angle whose measure is more than 180º but less than 360º

Angle around a point: An angle from the top of a ray to its bottom, completely

encompassing the defined point of the ray is called angle around a point. By definition

its measure is equal to 360º

Example 1:

AB is a line and C is a point on it D is a point away from the line.

If A C D = 48ofind Angle B C D

A C D +B C D = 180º Angles on a line are supplementary

B C D =180º – A C D = 180º -48º = 132º

Example 2:

AB and CD are two lines intersecting of point O. If angle AOC is equal to 50º

Find angle COB, BOD and AOD

A D

50º O

C B

A O C = B O D =50º (vertically opposite angles)

A O C+B O C = 180º (Angles on a line)

B O C =180º - A O C = 180º - 50º =130º

B O C = A O D =130º (vertically opposite angles)


Example 3:

AOB is a line C is a point away from the line. Angle AOC is equal to 4xº and angle BOC

is equal to 5xº Find the value of x.

C

4xº 5xº

A O B

4xº + 5xº =180º (Angles on a line)

9xº =180º

x = 20º

Example 4:

Three rays AO, BO and CO emerge from point O. Angle between any two adjacent rays

(lines) are equal to any another. Find the measure of the angle.

B

A O

C

A O B +B O C+C O A = 360º (Angles around a point)

But B O C = A O B and C O A = A O B

A O B+A O B+A O B = 3A O B = 360º A O B =120º

Example 5: PO and QO are line segments of opposite rays. RO and OS are two lines

emerging from O such that P O R =Q O S. Also R O S = 90º Find P O R and Q O S


R S

P O Q

P O R + R O S + S O Q = 180º -------- (Angles on a line)

P O R + 90º + S O Q =180º ----------(R O S = 90º)

P O R + 90º + P O R =180º ------- (P O R = S O Q)

2P O R = 90º or P O R = 45º = S O Q

Example 6: If A O B shown in the figure if 45º, find the measure of the corresponding

reflex angle.

A

O B

Interior A O B + Exterior A O B =360º ------- (Angle around a point)

Exterior A O B =360º – 45º =315º =Reflex angle

Example 7: Identify the following features in the figure shown below:

a. Opposite rays

b. Intersecting lines

c. Point of intersection

d. Opposite angles

e. Linear pair of angle


C

A O B

D

Workings

a. AO and OB, CO and OD

b. AB and CD

c. O

d. A O C and B O D, B O C and A O D

e. A O C and B O C, A O D and B O D

Example 8: AO and BO are two rays meeting at point O, Another ray OC bisects the

angle AOB such that A O C = B O C. If A O B measures 60º, find the values of A O C and

B O C

A

C

O B

A O B = A O C+B O C

A O B =2A O C

2A O C =60º

A O C =30º =B O C

Example 9:

Angles A O C and B O C are vertically opposite angles. Rays (lines OE and OF are

bisectors of angles A O C and B O C respectively. Show that EF is a line passing through

O.


A D

E O F

C B

A O D + D O F + F O B + B O C+C O E + E O A = 360º

(Angles around the point O)

But C O B = A O D ----- (Opposite angles)

C O E = A O E ---- (EO is the bisects or A O C)

And B O F = F O D. A O E = B O F

Hence EF is a line

Example 10:

AB is a line and O is a point on it, and C is a point away from it. P and Q are points such

that OP bisects A O C and OQ bisects B O C. Show that P O Q is a right angle.

C

P Q

A O B

Because PO and QO are angle bisectors

A O P = P O C

and C O Q = Q O B

A O P + P O C + C O Q + Q O B =180º

P O C + P O C + C O Q + C O Q =180º

2P O C + 2C O Q =180º

P O C + C O Q = 90º = P O Q


EXERCISES

1. State True or False

1. Two distinct points in a plane can form a part of straight line.

2. If two adjacent angles are equal that each has to measure 90º.

3. Two lines in a plane can have more than one point in common.

4. Angles forming a line can be both obtuse angles

5. Two adjacent angles equal in measure, adding to 180º are right angles

6. If two lines intersect vertically opposite angles are equal

7. Angles around a point add up to 180º

8. If two angles add up to 180º they are called supplementary angles

9. A point has no size

10. An angle bisector divides an angle into three equal parts.

2. O is a point on the line AB, CO is a ray drawn such that A O C =90º, find the

measure of angle B O C

3. In the following figure A O C =3xº and B O C =7xo find the value of x

o

C

3xº 7xº

A O B

4. If PO, QO and RO are three rays from the points O. If P O R =60º and Q O R

=120º show that PO and QO are opposite rays.

5. In the following figure A O C is greater than B O C by 30º, find the value of both

the angles

C

A O B


6. 3 rays AO, BO and CO emerge from a point ‘O’ such that A O B = B O C = C O A

=120º three more rays PO, QO and RO are bisectors of the angles. Show that

P O Q, Q O R and R O P are also equal to 120º also show that AO and QO are

opposite rays.

A

R P

O

C B

Q

7. Lines AB and PQ intersect at point O if A O P =90º find P O B, B O Q and Q O A

8. In the figure shown below find the value of x please note the AB is a line and O is

a point on it.

2x

x + 10 x

A O B

9. AB is a line and O is a point one it and R is a point away from it. It A O R =8x and

B O R =10x. Find the value of x and the angles A O R and B O R.

10. AB and CD are lines intersect at point E line EF bisects angle AEC. If angle AEF

is 30º find all the remaining angles shown in the figure.


A

D

30º 1

2

F 4 E

3

C B

PARALLEL LINES

Two lines are said to be parallel to each other, if they do not intersect any where and are

in the same plane. Separation of two parallel lines is equal at all points. A transversal is

a line that intersects both the parallel liner. Several properties of parallel lines can be

explained with the help of a transversal. The diagram below shows a set parallel lines

with a transversal.

E

4 1

A B

3 2

8 5

C D

7 6

F

Description:

AB and CD – Parallel lines

EF –Transversal

1, 5, 2, 6, 3, 7 and 4, 8 pairs of corresponding angles

2, 5 and 3, 5 –Alternate angles

3, 8 and 2, 5 –Co interior angles

1, 3, 2, 4, 5, 7 and 6, 8 – Opposite angles.

PROPERTY- Corresponding angles are equal. The property when a pair of parallel lines

are intersected by a transversal the corresponding angles are equal is assumed to be true.

Hence this property is called the corresponding angles axiom. The converse is also true.

When the corresponding angles formed by a transversal are equal, the two lines it

intersects are parallel.


a

b

Corresponding angles a = b

Theorem

If a transversal intersects two parallel lines then alternate angles are equal

4 1

3 2

8 5

7 6

Working: In the diagram shown above angles 3 and 5 are a pair of alternate angles 1 and

5 and 3 and 7 are pairs of corresponding angles.

3 = 1 (vertically opposite angles)

But 1 = 5 (corresponding angles)

3 = 5

Theorem

If a transversal intersects two parallel lines then each pair of co interior angles are

supplementary.


4 1

3 2

8 5

7 6

Working: In the diagram above angles 2 and 5 are a pair of co interior angles and 1 and 5

are a pair of corresponding angles.

1+ 2 =180º ------ (angles or a line)

But 1 = 5 ------ (corresponding angles)

Hence 5 + 2 = 180º

Theorem

A transversal intersects two parallel lines if the alternate angles are equal then the two

lines are parallel

4 1

3 2

8 5

7 6

Working: 1 = 3 ------ (vertically opposite angles)

But 3 = 5 ------- (Alternate angles)

1 = 5, because the angles 1 and 5 are corresponding, as per the axiom the two

lines have to be parallel.

Theorem

A transversal intersects two lines. The co interior angles add up to 180º. In such a case

the two lines are said to be parallel


4 1

3 2

3 2

8 5

7 6

Working: 1+ 2 = 180º ------ (Angles on a line)

2+ 5 =180º ------- (Co interior angles)

5 = 1

Since 5 and 1 are corresponding angles, as per the theorem, the two lines are

parallel

EXAMPLES AND PROPERTIES

Example 1:

EF is a transversal cutting lines AB and CD at points P and Q respectively. If the angle

EPB (marked 1) is 30º, find all the remaining angles (2 to 8) marked. Give reasons

E

P 30º

A 4 1 B

3 2

8 5

C 7 6 D

Q

F

1+ 2=180º----- (Angle on a line)

30o + 2=180º ------ 2=150º = B P Q

1 = 3 =30º------ (Vertically opposite angles)

2 = 4 =150º ----- (Vertically opposite angles)

1 = 5 =30º------ (Corresponding angles)

2 = 6 =150º ----- (Corresponding angles)




Example 2: AB and CD are two lines cut by a transversal EF, intersecting AB and CD at

G and H. If EGB =40º = CHF show that lines AB and CD are parallel.

E

A G 40º B

H

C 40º D

F

AGF = EGB = 40º ---- (Vertically opposite angle)

Therefore AGF = CHF = 40º (It is CHF = 40º

But angles AGF and CDF are corresponding angles formed by a transversal intersecting

two lines. Hence by the axiom of corresponding angles being equal, the lines AB and CD

are parallel.

Example 3: PQ and RS are two lines intersected by a transversal TU at points X and Y

Angles PXY and TYS are equal prove that PQ is parallel to RS

T

X

P Q

R S

Y

U

PXY = TXQ ------- (Opposite angles)

But angle PXY = TYS ----- (As per the problem)

Hence TXQ = TYS


But angles TXQ and TYS are a pair of corresponding angles formed by a transversal

intersecting two lines. By the axiom, if corresponding angles are equal the two lines

intersected by the transversal are parallel.

Hence PQ is parallel to RS.

Example 4: A pair of parallel lines AB and CD are intersected by a transversal EF which

is also perpendicular to line AB. Show that EF is also perpendicular to CD.

E

A G B

H

C D

F

Let EF intersect lines AB and CD at G and H

E G B = E H D ---- (Corresponding angles)

But E G B = 90º

E H D = 90º

Hence EH is perpendicular to CD

Example 5: A transversal EF intersects a pair of parallel lines AB and CD. G and H are

points of intersection of the transversal and the parallel lines AB and CD respectively. If

AGE =33º find GHD

E

33º

A B

G

H

B D

F

AGE = HGB ------ (Opposite angles)

HGB = 33º ----- (AGE = 33º)


But HGB +GHD = 180º ----- (co interior angles)

33º +GHD = 180º

GHD = 147º

Example 6: AB and CD are a pair of parallel lines intersected by a transversal EF. G and

H are points of intersection. GI and HK are bisectors of angles EGA and DHF

respectively. Prove that GI is parallel to HK.

E

G

A B

K

I

C H D

F

AGH = GHD ------ (Alternate angles)

ButAGH = 2 IGH ------- (IG is the bisector of AGH)

Also GHD = 2GHK ----- (HK is the bisector of GHD)

2 IGH = 2GHK

or IGH = GHK

IGH and GHK are alternate angles formed by transversal GH intersecting the lines

IG and HK. They being equal IG must be parallel to HK.

Example 7: PQ and RS are two parallel lines MN is a transversal line perpendicular to

both PQ and RS and intersecting them at t and u from a point ‘V’ on the line PQ a line Vu

is drawn so that vu bisects MUR. Show that TVU = 45º

M

V T

P Q

R S

U

N


MU is perpendicular to RS

MUR = 90º

VU is the angle bisector

VUT =902 = 45º = VUR

But Vu is also a transversal to the lines PQ and RS

TVU= VUR = 45º (Alternate angles)

Example 8: AB and CD are two parallel lines a transversal. MN intersects both of them at

points P and Q. MN also intersects another line EF at point R. If APQ = ERN, show

that CD is also parallel to EF.

M

A P B

C Q D

E R F

N

APQ = CQR ------ (Corresponding angles)

But APQ = ERN (as given in the problem)

ERN = CQR

But ERN and CQR are corresponding angles formed when the transversal MN

intersects lines CD and EF. By the axiom of corresponding angles we have that CD is

parallel to EF.

Example 9:

In the figure shown below lines AB and BC originate from point B and lines DE and

EF originate from point F. Also AB is parallel to DE and BC is parallel to EF. Points B

and E have been joined and the line EB has been extended to the point G. If BG angle

ABC prove that EB bisects angle DEF.

D A

B

E G

F C


EG is a transversal to both the pairs of parallel lines AB, ED and DE, EF.

DEB = ABG ------- (Corresponding angles)

But ABG = CBG =½ ABC (BG is the bisector or ABC)

DEB = ½ ABC

like wise FEB = CBG ---- (Corresponding angles)

But CBG = ABG = ½ ABC ----- BG is bisector of ABC)

CBG = FEB = ½ ABC

FEB = ½ ABC = DEB

Hence EB is the bisector of DEF

Example 10:

In the figure shown below AB, CD and EF are parallel lines MN is a transversal

intersecting lines AB and CD. NP is a transversal intersecting lines CD and EF. If

BMN = 70º andFPN = 140º Find MNP.

A M B

70º

C N D

140º

E F

P

BMN +MND = 180º----- (co interior angles)

MND = 180º - BMN = 180º– 70º

MND = 110º

EPN +PND = 180º ------- (co interior angles)

PND = 180º- FPN = 180º – 140º

PND = 40º

MNP = MND + NPE ------ (As per Diagram)

(adjacent angles)

MNP = 110º+ 40º

MNP = 150º

EXERCISES

1. Which of the following are True and False

a. If two parallel lines are intersected by a transversal the co interior angles are

equal

b. If two parallel lines are intersected by a transversal then the corresponding

angles are equal

c. Parallel lines will intersect at two points.


d. Two lines perpendicular to the same line are perpendicular to each other.

e. Two lines parallel to the same line are not parallel to each other.

f. Co interior angles are on the opposite side of the transversal

g. Angle bisectors of alternate angles are also parallel

h. Two transversals of a set of parallel lines never meet.

i. Converse of the corresponding are equal if the transversal intersects two

parallel lines is not true.

2. Name the following refer to the diagram below.

E

A G B

C H D

F

a. Pair of parallel lines

b. Transversal

c. Angle corresponding to AGE

d. Co interior angle pairs

e. Alternate angle to AGH

3. PQ and RS are a pair of parallel lines OP is a transversal intersecting PQ and RS

at M and N respectively. If PMO is 40º Find angles, NMQ, PNR, PMN and OMQ

O

M

P Q

N

R S

P

4. QR is the transversal perpendicular to both the lines MN and OP which it

intersects. Prove that MN and OP are parallel to each other.


5. In the figure below two parallel lines are intersected by two transversals which are

parallel to each other. If the angle marked measure 55º find the value of angle

marked 2, 3, 4 and 5

1 2

5 4 3

6. In the following figure QRS = PQR =70º STU and RST = 150º

If PQ ׀ RS show that RS ׀ ׀ ׀ TU ׀ ׀ PQ

P Q

70º T U

150º

70º 150º

R S

7. In the figure below bisectors GO and HO of co interior angles BGH and GHD

meet at a point O. Show that GOH is a right angle

E

A B

G

O

H

C D

F


8. If two lines are intersected by a transversal such that the co interior angles formed

on one side add up to 180º show that the two lines are parallel.

9. In the following figure identify the pair of parallel lines, state the reason.

A C

100º 100º

P Q

120º

R 120º S

B P

10. Lines PQ and RS are parallel. T is any point between the lines. An angle QTS is

formed by joining QT and TS. Show that PQT + RST = QTS given

QTS is less than 180º

P Q

T

R S


TRIANGLES

A triangle is the simplest geometrical figure obtained by 3 lines. A figure so formed

contains 3 interior angles and hence called a triangle. They can be classified on the basis

of sides or angles.

1. Scalene Triangle:

A triangle with all the three sides having different length is called a scalene triangle.

2. Isosceles Triangle:

A triangle with two sides having the same length is called an isosceles triangle.

3. Equilateral Triangle:

A triangle with all the three sides having the same length is called an equilateral

triangle.


4. Acute angle triangle:

A triangle with all the three angles acute is called an acute angle triangle.

5. Right angle triangle:

A triangle with one of its angle as a right angle is called a right angled triangle.

6. Obtuse angle triangle:

A triangle with one of its angle more than 90º is called an obtuse angle triangle.

Properties:

A triangle is usually designated by three letters attached to the vertices. For example

ABC indicates a triangle whose vertices are A, B and C also A, B and C denote the

interior angles of a triangle. Length of sides opposite A, B and C are indicated by small

care letter a and c. However the side joining any two vertices is given as AB, BC and CA.

In a triangle the sum of length of any two sides will be more than the length of the

remaining side.

a + b > c

or BC + CA > AB

A

c b

B C

a


In a triangle the sum of all the length of sides is called the perimeter. Also the side

opposite to the largest angle will be the biggest. Three angles of a triangle always add up

to 180º and an exterior angle is equal to the sum of two opposite interior angles. Proof

and explanation of these two angle properties are given in the following theorems.

Theorem: Three interior angles of a triangle add up to 180º

A

D E

Consider a ABC as shown in figure.

B C

Draw a line parallel to BC through the point A then AB and BC become transversals

Hence ABC = BAD ------ (Alternate angles)

ACB = CAE ----- (Alternate angles)

But BAD + BAC + CAE = 180º (Angles on a straight line)

ABC +BAC + CAE = 180º

Hence angles in a triangle add up to 180º

Theorem: If one of the sides is extended so as to form an exterior angle then the exterior

angle will be equal to the sum of opposite interior angles.

A

B C D

In the triangle ABC line BC is extended to D

ABC +BCD = 180º ----- (Angles on a straight line)

But inABC

ABC + BAC +BCA = 180º

ABC +BAC + ACB = ACB +BCD

ABC +BAC = BCD (Exterior angle)

Hence an exterior angle is equal to the sum of opposite interior angles.

Example 1: Identify which of the following figures belong to a triangle

(1) a = 15cms, b = 5cms, c = 6cms

Ans: ABC is not a triangle because a is greater sum of b and c


(2) A = 90º, B = 60º and C = 30º

Ans: A +B +C = 180º, hence ABC is a triangle

(3) Exterior angle ACD = 100º, Opposite interior angles CBA and BAC are 50º and 30º

Ans: ACD = 100º, and CBA +BCA = 80º only and not 100º Hence ABC is not a

triangle

(4) a = 10, b = 8 and c = 6

Ans: a = 10 is less than b + c = 8+ 6 =14 hence ABC is a triangle

(5) A = 70º, B = 60º and C = 40º

Ans: A + B + C = 70º + 60º + 40º = 170º and is not equal to 180º, Hence ABC is

a triangle

Example 2: In the triangle shown below A = 60º and B = 80º Find C

A

60º

80º

C B

Example 3: In a triangle shown below exterior angle ACD = 105º and BAC = 45º.

Find the remaining angles.

A

45º

105º

B C D

ACD = BAC + ABC

105º = 45º +ABC

ABC = 60º

Also ABC + BAC + ACB = 180º

60º + 45º + ACB = 180º

ACB = 75 º


Example 4: In an isosceles triangle A = 80º, B = 50º and C = 50º. If the side

opposite A is 8cm = a and side b opposite to B = 10.3cms. Find the perimeter of the

triangle.

A

80º

10.3cm

B 8cms C

Ans: In the triangle ABC

B = C = 50º

AB = AC

AB = 10.3cm = BC

Perimeter = AB + BC +AC

= 10.3 + 8 +10.3

= 28.6cms.

Example 5: In an equilateral triangle ABC show that each of the angle is equal. If the

perimeter of the triangle is 15cms. Find the length of each side and each angle.

A

B C

A + B + C =180º ---- (Angles in a triangle)

But A = B = C ----- (ABC is an equilateral triangle)

A + A + A = 180º or 3 A =180º A = 60º= B = C

a + b + c = 15cms

But a = b = c a + a + a = 15cms

Or 3a = 15cms

a = b = c = 5cms


Example 6: PQR is an isosceles triangle with PQ = QR side QP is extended to S such

that QP = PS show that SRQ = 90º

S

P

Q R

Ans: QPR = PSR +PRS ---- (Exterior angle theorem)

But ∆PSR is also Isosceles because PS = PR

PSR = PRS

Hence QPR = 2 PRS

Similarly

SPR = PQR + PRQ ------ (Exterior angle theorem)

But ∆ PQR is Isosceles

PQR = PRQ

SPR = 2 PRQ

QPR + SPR = 2 PRS + 2 PRQ

But QPR + SPR = 180º ------ (Angles on a line)

2 PRS + 2 PRQ = 180º

PRS + PRQ = 90º

But PRS +PRQ = QRS ------- (Adjacent angles)

QRS = 90º

Example 7: ABC is triangle and AD is the angle bisector. If ABC = 60º and ACB = 40º,

find all the remaining angles.

A

1 2

60º 3 4 40º

B D C


Angles marked

1 = BAD

2 = DAC

3 = ADB

4 = ADC

BAC + ABC + ACB = 180º ------ (Angles in a triangle)

BAC + 60º + 40º = 180º

BAC = 80º

But AC bisects BAC

BAD + DAC = BAD + BAD = 80º

or 2 BAD = 80º or BAD = 40º

BAD = DAC = 40º (AD is the angle bisector)

ADC is the exterior angle to ∆ ABC

ADC = ABD + DAB

ADC = 60º + 40º

ADC = 100º

But ADB + ADC = 180º --------- (Angles on a straight line)

ADB + 100º = 180º

ADB = 80º

Example 8: If the base of an isosceles triangle extended on both sides show that the

exterior angles so formed are also equal

P

S Q R T

Ans: Consider a triangle PQR with base QR extended to S and T on both sides

PQR = PRQ ----- (PQ = PR and PQR is Isosceles)

180º – P Q R = 180º – P R Q

But 180º – P Q R = P Q S ---- (Angles on a straight line)

And 180º – P R Q = P R T

P Q S = P R T


Example 9: PQRS is a Quadrilateral and PR is a diagonal. Show that the sum of interior

angles add up to 360º

P

S

Q

R

Ans: Consider ∆PSR PRS + RPS + PSR = 180º ------- (Angles in a triangle)

and QPR + QRP + PQR = 180º ------- (Angles in a triangle)

PRS + RPS + PSR = 180º ---- (1) (Adding equations 1 & 2)

*QRP + QPR + PQR = 180º ----- (2) (and combining adjacent angles)

QRS + QPS +PSR + PQR= 360º ----- (Angles in a Quadrilateral)

Example 10: ABC is a triangle with side BA extended to D, AC to E and CD to F. Show

that the exterior angles formed add up to 360º

D

A

C

F B

E

FBA +ABC = 180º ---- (1) ------- (Angles on a straight line)

DAC +BAC = 180º ---- (2) ---- (Angles on a straight line)

ECB + ACB = 180º ---- (3) ---- (Angles on a straight line)

Adding equations (1), (2) and (3)

FBA +DAC + ECB +BAC + ACB+ABC = 180º + 180º + 180º

FBA + DAC +ECB + 180º = 180º + 180º + 180º

Because ABC + BAC +ACB are interior angles of a triangle and hence add up to

180º

FBA + DAC + ECB = 360º



1. Three angles of a triangle add up to 180º

2. An obtuse angled triangle can have two interior obtuse angles.

3. Each one of the interior angles of an equilateral triangle measures 60º

4. ABC is a triangle with side AB measuring 12cms BC = 5cms and CA = 6cms

5. PQR is a triangle with P = 80º, Q = 60º and R = 40º

6. In an equilateral triangle any two sides will be equal to each other.

7. In an Isosceles triangle sides opposite to equal angles are equal

8. A triangle is not polygon

9. In a triangle the side opposite to the largest angle is the smallest

10. In a triangle the side opposite A is denoted as ‘a’

2. In a triangle ABC, B = 2 A and C = 3 A Find the values of all the angles.

3. In a right angled triangle PQR show that one of the angles is equal to the sum of

remaining two angles.

4. ABC is scalene triangle BO and OC are bisectors of the interior angles CB ˆ and ˆ

meeting at point O. Prove that 2

ˆˆ180o CB

.

5. In the triangle ABC show below, B A C is 40º and the exterior angle CAD is 100º

Find the remaining two angles.

A

40º

100º

B C D

6. In the figure shown below AB = AC = CD and BAC = 50º Find all the values of

remaining interior angles.

A

50º

B C D


7. In the triangle PQR, P = 40º and Q = 60º Identify the largest and smallest sides

8. In an Isosceles ∆ABC side AB = AC show that the bisector of exterior angle CAD

is parallel to base BC.

D

A

B C

9. Calculate the values of angle x, y and z in the following figure

x

110º y z 130º

10. In the regular pentagon shown below find the sum of all the interior angles

B

A C

E D


CONGRUENCY OR TRIANGLES

Any two figures which are identical, that is having the same shape and size are said to be

congruent. To prove congruence of triangles is an important step in learning plane

geometry. Two figures are congruent, if one figure can be super imposed on another so

that they cover each other exactly.

Simplest of the figures a line segment is said to be congruent if they are of equal length.

Two angles are said to be congruent if both of them have the same measure. Hence, two

triangles are said to be congruent if all the three sides and three angles are equal.

Therefore all parts namely sides and angles have to be equal to prove that the triangles

are congruent.

A P

B C Q R

Congruent triangles

PQRABCPRQ ACB QPRBAC

QR, PR, BCACPQ,ABABC

and,

ifwith congruent is above figure In the

However, a triangle can be constructed if we know two sides and an angle or two angles

and a side or all the three sides. Only three out of six parts of a triangle need to be known

because the remaining three parts can be calculated. But it is important to note that two

triangles can not be said to be congruent if only the three angles are found to be equal. If

only the angles are equal the two triangles, will have the same shape and not necessarily

the same size.

Some of the important properties and theorems regarding congruency of triangles have

been detailed below.

Property 9.1:

If ABC is congruent to DEF then DEF is congruent to ABC . Congruency is

represented as ‘ ’ in the Geometric expressions

PQRABC

PDEFDEFABC

ABCDEFDEFABC

then

QRΔ and if Also

Δ then If

Theorem:

Prove that two triangles are congruent if two angles and the included side of one triangle

is equal to the two angles and included side of the other.


DEFABC

DA

DFAC DEAB

AH

Hence

ˆˆ Also

and Therefore

with coincide should Hence

If CA, BDEAB and points will exactly coincide with DEF Therefore all the angles

and sides of both the triangles are equal

DEFABC

Reconstruct DEF such that EF is superimposed and exactly coincides with DBC and is on

the opposite side of ADBC join

ABC is an isosceles triangle because

(Given that sides ABC are equal to sides of DEF )

ABD BDABAD (Δ------- is isosceles)

A D

H G

B C E F

In the triangles ABC and DEF

B = E, C = F and BC = EF

If we superimpose ∆ABC over ∆DEF we have three possibilities

(a) AB = DE or (b) AB < DE or (c) AB > DE

If AB < DE then let ∆EGF be the super imposed position of ∆ABC

In the fig HCB =ACB EFD = GFD which cannot be true if G is not the same

as D.

Hence point G should co inside with point D

Therefore we have AB = DE and DF = AC

Also A + B + C = 180º = E + F + G

A = E and B = F

We have C = G

Hence ∆ABC ∆DEF

The same argument can be extended to prove ∆ABC ∆DEF if AB > DE also.

Theorem: Two triangles are congruent if two sides and their included angles are equal.


Consider two triangles ABC and DFE having sides AB = DE and BC = EF and ABC =

DEF

A D

B C E F

A (D)

B (E) C (F)

Draw line BC and reconstruct ∆ ABC such that vertex E coincides with vertex B

DEF = DEF ------- (E coincides with point C)

And line EF coincide with line BC, because BC = EF

Because ABC has the same measure of DEF line DE will coincide with line DE.

Also AB and DE are if equal length and as such vertex A will coincide with vertex D.

Also line BC coincides with line EF, as per construction. Because BC = EF, F will

coincide with C.

Therefore line AC should coincide with line DF

AC = DF

Hence, AB = DE, BC = EF and AC = DF

∆ ABC ∆ DEF

Example 1: ABC is an isosceles triangle AD is perpendicular to BC and D lies on BC.

Show that ∆ ADB and ∆ ADC are congruent.

A

B D C


Ans: If AD is BC

ADB = ADC = 90º

Also AB = AC ------ (∆ ABC is isosceles)

And ABC = ACD ------ (∆ ABC is isosceles)

Since corresponding two angles and a side are equal in triangle ADB and ADC

∆ADB ∆ADC ------- (ASA Condition)

Example 2: ABC and BCD are two equilateral triangles with a common base BC vertices

A and D are joined by a line AD. AD intersects BC at point E. Show that AD and BC are

a pair of perpendicular bisectors.

A

B E C

D

Ans: Consider ∆ABD

AB = BD ------- (∆ABC ∆DBC)

BAD = BDA ------ (∆ABD is isosceles)

Consider ∆ACD

AC = CD ------ (∆ABC ∆DBC)

CAD = CDA ------ (∆CAD is isosceles)

∆ABC is equilateral

AB = BC = AC

And ∆BCD is equilateral

AC = BD = CD

Hence AB = AC = BC = BD = CD ------ (BC is common side)

∆ ABC is equilateral

ABC =BAC = BCA = 60º

∆ BDC is equilateral

BDC = BCD = DBC = 60º

ABC = BAC = BCA = BDC = BCD = DBC = 60º

Now consider triangles BAE and EDC

BA = DC

ABC = ABE = ECD = BCD = 60º

AEB = DEC --------------- (vertically opposite angles)

∆ BAC ∆ EDC ------------- (ASA criteria)

AE = ED


and BE = EC

Hence AD and BC are a pair of bisectors intersecting of point E

Consider triangles ABE and DBE

ABE =DBE = 60º

BAE = BDE - - - - - - - (∆ BAD is isosceles)

The remaining pair of corresponding angles

BEA = BED

But BEA + BED = 180º------------ (Angles on a line)

2BEA = 180º

BEA = 90º

= BED

Hence BE and ED or BC and AD are per perpendicular

Example 3: ∆ABC and ∆PQR are congruent scalene triangles D and E are mid points of

AB and AC. S and T are mid points of PQ and PR. Show that ∆ADE is congruent to

∆PST

A P

D E S T

B C Q R

AB = PQ -------------- (∆ABC ∆PQR)

But AB = 2AC and PQ = 2PR ----------- (D and S are mid points)

2AD = 2PR

AD = PS

AC = PR -------------- (∆ABC ∆PQR)

But AC = 2AE and PR = 2PT --------- (E and T are midpoints)

2AE = 2PT

AE = PT

Consider the included angle DAE is the same as BAC and SPT is the same as

QPR

But BAC = QPR -------------- (∆ABC ∆PQR)

DAE = SPT

Since two sides and their included angle are identical in ∆DAE and ∆SPT

∆DAE ∆SPT


Example 4: In two triangles ABC and DEF the exterior angles and interior angles at

points B and E are equal. Also side BC is equal to side EF. Are the triangles congruent?

Explain your answer.

No, the exterior angle is equal to sum of the opposite angles. Hence the interior angles of

∆ABC may not be equal to those of ∆DEF.

Example 5: ∆ABC and ∆PQR are two scalene triangles. BC is extended to D such that

BC = CD and QR is extended to S such that QR = RS given ∆ABC is congruent to ∆PQR

show that ∆ABD is also congruent to ∆PQS

A P

B C D Q R S

BD = 2BC ----------- (Given BC = CD)

And QS = 2QR ----------- (Given QR = RS)

But BC = QR ------------- (∆ABC ∆PQR)

BD = QS

Also AB = PQ -------------- (∆ABC ∆PQR)

ABC = ABD = PQR = PQS ------------- (∆ABC ∆PQR)

Two corresponding sides and included angles are equal in triangles ABD and PQS

∆ABD ∆PQS -------------- (SAS criteria)

Example 6: ABC and PQR are two triangles in which (AB +BC) = 7cm and CA = 5cms.

Lengths of all the sides are integers. Also

PQ = 3cm and (PR + QR) = 9cm

Check if ∆ABC, ∆PQR if they are both right angled triangles.

Check (AB+BC) +CA = PQ + (QR +PR) =

7 + 5 = 3 + 9

Assuming the triangles are congruent

(QR +PR) – (AB +BC)

(AC + BC) – (AB +BC) = AC – AB = 9-7 =2 - - - - - - - - (Assumed AC = QR and BC =

PR)

But AC = 5cms

AB = 5-2 = 3cm and AB + BC = 7cm BC = 4cm

AB = PQ = 3cm

AC = PR = 5cm

BC = QR = 4cm


Triangles with sizes of sides mentioned above can be verified to be right angled triangles

as given from Pythagoras theorem

32 +4

2 =5

2

Hence no other combination of side length is possible

∆ABC ∆PQR ----------- (S.S.S criteria)

Example 7: ABCD is a square AC and BD are two diagonals. Prove that ∆ABC is

congruent to ∆DCB Hence show that AC = BD

A B

D C

AB = DC ------------------- (ABCD is a square)

And AD = BC ---------------- (ABCD is a square)

BAC = BCD ---------------- (ABCD is a square)

Since two corresponding sides and included angles are equal ∆ADC ∆BDC

The remaining pair of corresponding sides

AC = BD

Example 8: ABC and DEF are two triangles. Congruent triangles as shown below. Name

all the corresponding parts and find their value.

A D

70º 5cm 4cm 70º

60º 50º

B 6cm C E 6cm F

Corresponding parts ∆ ABC ∆ DEF Remarks

Side AB – side DE 4cms 4cms AB = DE = 4

Side BC –side EF 6cms 6cms Given

Side AC – side DE 5cms 5cms DF = AC = 5


ooo

ooo

oo

50 C B 50 50 FC

60 E B 60 60 EB

Given 70 70 DA

Example 9: Show that two right angled triangles are congruent if the hypotenuse and a

side are equal.

Working: Let ABC and DEF be two right angled triangles

A A (D)

B C B C (E) F

Given Hypotenuse AB = DF

Side AC = DE o90ˆˆ EC

Reconstruct ∆DEF such that AC coincides with DE and F is on the other side of B

Now BCA =ACF = 90º ------------- (DEF and ACF are same because DE

coincides with AC)

BCA + ACF = BCF

90º + 90º = BCF

or BCF = 180º

B, C and F are on the same line

Hence ABF is a triangle

AB = AF = (DF) ABF is isosceles

ABC = AFC = DFC

Hence in triangle ABC and DEF we have two angles and the included side are equal

∆ABC ∆DEF

Example 10: Two lines AB and CD bisect each other at S. Prove that ∆ADS is

congruent to ∆CSB

A D

S

C B


S is the mid point of AB

AS = SB

S is also the mid point of CD

CS = SD

Also ASD = CSD ------------------ (Vertically opposite angles)

Since two sides and the included angle are equal

∆ASD = ∆CSB

Example 11: AB and CD are a pair of parallel lines cut by a transversal MN at points P

and Q. Bisectors of APQ and CQP meet at R and bisectors of BPQ and DQP

meet at S. Prove that ∆PRQ is congruent to ∆PSQ

Working: AB׀׀CD and PQ is a transversal

APQ = DQP ---------------- (alternate angles)

BPQ = DQP ---------------- (alternate angles)

M

P

A B

R S

C Q D

N

Since PR bisects APQ and SQ bisects DQP

APR =2

1 APQ =2

1 DQP = PQS

since QR bisects CQP and PS bisects BPQ

PQR =2

1 CQP = 2

1 BPQ = 2

1 SPQ

Also side PQ is common to both the triangles

PQR and PQS

Hence ∆PQR ∆PQS -------------- (ASA criteria)

EXERCISES

1. ABC and DEF are two isosceles triangles prove that ∆ABC ∆DEF if any one

side and an angle are equal.

2. AX and BY are two lines of equal length, drawn from the end points A and B of

line AB. AX and BY are parallel and are on the opposite sides of line AB. XY is

joined and intersects AB at Z. Show ∆AXZ BYZ


X

Z

A B

Y

3. ABC and DEF are two congruent triangles P is the midpoint of BC and Q is the

mid point of EF given BC = EF. Show that ∆APB ∆DQE and ∆APC ∆DQF.

4. ∆ABC is congruent to ∆PQR List all the criteria to be satisfied for the triangles to

be congruent. List all the combination of corresponding sides and angles to be

equal for the triangles to be congruent.

5. ABC is an isosceles triangles with B =C. CD and BE are perpendicular to

lines BC and AB respectively. Prove that ∆BCD is congruent to ∆BCE and hence

show that BE = CD.

6. ABC and DEF are two triangles having a perimeter of 12cms each. If the longest

sides AB and DE are 5cms each and side BC is 4cms where as side DF is 3cms.

Show that both the triangles are congruent.

7. ABC is an equilateral triangle D, E and F are mid points of AB, BC and CA.

Show that ∆ADF ∆BDE ∆CEF ∆DEF.

A

D F

B C

E

8. AB and CD, PQ and RS are two sets of perpendicular lines. AB meets CD at

point E and PQ meets RS at T. Also AE = PT, and DE = RT. Show that two

triangles ACD and PRS are congruent.


C R

E T

A B P Q

D S

9. ABC and DEF are two congruent triangles placed such that B, C, E and F are on

the same line. Show that AB is parallel to DE and AC is parallel to EF. AD

when joined will be parallel to BC or EF.

A D

B C E F


LOCI AND CONCURRENCY

Loci:

Locus of a point is the path traced by it under some given mathematical conditions.

Plural of locus is loci, simplest of the examples of a locus is a circle. A circle is traced by

a moving point which moves in such a way that it i.e., at constant distance from a fixed

point called the center of the circle. In terms of practical Geometry the fixed point is the

sharp point of the compass and the moving point is the sharp end of the pencil.

O r

Locus of a line at equidistant from another line is a parallel line. Converse of a loci is to

find the mathematical relationship given the locus and the reference point or a feature.

There are many practical application for the study of loci. For example trace a specific

path of a planet around the sun with the sum as the reference point. Study of such path

helps in predicting their location at any time.

Theorem: Locus of a point which is at an equal distance form two fixed points is the

perpendicular bisector of the line joining the fixed points.

Let C, D be the positions of moving point and A B the fixed points.

C

A D B

Let D be the position of the moving point

When it is on the line AB

Then it is given that AD = DB and hence

D is the mid point of line AB

Consider another point on the locus away from D

Then it is given AC = CB

Hence CAB =CBA

In triangles ACD and ADB

CAB = CBA


AC = DB

and AD = DB

∆ADC ∆BDC -------------- (SAS criterion)

CDA = CDB

But CDA + CDB = 180º ---------------- (Angles on a line)

2CDA = 180º CDA = 90º = CDB

Since AD = AB, CD is the perpendicular bisector of AB and the locus is the

perpendicular bisector

Theorem: Prove that the locus of a point equidistant from two lines meeting at a point is

angle bisector of the angle formed by the two lines

A

P

B C

Let BP is the locus of the point

Then AP = PC -------------- (Given)

Since AP and CP are the shortest distance form P to line AB and AC

BAP = 90º = PCB ------------ (perpendicular is the shortest distance form a point to a

line)

ABP = PBC, Hence PB is the bisector ABC

∆BAP and ∆BCP are right angled triangles form Pythagoras theorem we have

BP2 = BA

2 + AP

2

ABC ofbisector theis BP hence BPC ABP

BDC ABP BPC, of sides three the toequal is ABP sides threeSince

BC PB - BP PA - BP AB

PB PA

PB - BP BC

BP BC BP also

PA-BP ABor

2222

22

222

22

Example 1: What is the locus of a point P always at a distance of 1.5cms from a fixed

point O. Draw the locus.

O

1.5cms

P


Working: The locus P will be a circle of radius 1.5cms and center O- Figure 10.1.1

shows the locus of P and a point on the locus P and the center O

Example 2: A and B are two fixed points on a line CD. P and Q are two points on the

loci moving such that PAB = 60º = QBD. Show that the Loci PA and QB are

parallel.

P Q

C A B D

Working:

PAB = QBD = 60º --------------- (Given)

But PA and QBD are corresponding angles

PA׀׀QB ---------------- (corresponding angles axiom)

This axiom is true of all length of PA or QB which

represent the locus of P and Q

Example 3: Show that the locus of the vertices formed by a moving point P with a fixed

base QR of isosceles triangles is the perpendicular bisector of QR, lets be the point of the

locus on QR.

PQ = QR

PQR = PRQ or

∆PQS ∆PRS

QS = RS

Also PSQ = PSR = 90o

Hence the locus is the perpendicular bisector

P1

P2

Q R

S

P3

Working: Because P1QR is isosceles

PQ = QR


P1QR = P1RQ or P1QS = P1RS

Also P2QR is isosceles

criteria) (SAS ------------ RPPQPP

common is PP

RPP QPP

RP QP

have weRPP and QPP sin tringle Also

angles)(Adjacent ---------RPP PQP

RSP - RSP QSP - QSP

RSP QSP

21 21

21

21 21

11

2121

212

2121

22

SR SQ and

SRP 90 SQP

angles)(Adjacent ----------- 180 SRP SQPBut

SRP SQP

RSP QSP

common is SP

SRP SQP

RP QP

RSP and QSP esIn trianlg

SRP SQP

line) aon (Angles 180 SRP RP P SQP QPPBut

RRP QPP

2

o

2

o

22

22

22

2

22

22

22

22

o

221221

2121

hence P2S is the perpendicular bisector. This result can be shown for any point P1, P3 or

any other vertex of the isosceles triangle formed by base QR. Hence the perpendicular

bisector of QR is the locus.

Example 4: Find the locus of the mid points of the parallel chords of a circle.

A B

P2

C P1 D

E F

O


Working: Let EF be the diameter (longest chord) of a circle passing through the center O

let AB be a chord parallel to EF with mid point P2 and CD be another chord parallel to

EF with mid point P1.

Consider triangles OP2B and OP2A

OA = OB ----------- (radii of a circle)

AP2 = P2B ---------- (Given P2 as mid point of AB)

OP2 is common

o

22

o

22

22

22

90 BOP A OP

line) aon (Angles ----------- 180 BOP A OPBut

BOP A OP

A OP BOP

Since AB׀׀ CD and OP2 is a transversal cutting CD at P1

CP1O = DP1O = 90o - - - - - - - - - - - - (Corresponding angles)

Consider triangles CDO and DPO

OC = OD - - - - - - - - - - - - - (radii of a circle)

CP1O = DP1O

P1O is common

DP CP

DPO CPO

11

Hence P1 is the mid point of CD and P1O is perpendicular to CD and P1 is a point on the

line OP2. Hence the locus of midpoints of chords parallel is their perpendicular bisector

which is also a diameter of the circle.

Example 5. Find the locus of points which are equidistant from three given points.

These points are not in a line

A

R P

B C

S

A, B and C are three points AB and BC are joined. Let R be the mid point of

AB and S the mid point of BC.

Locus of a point equidistant form AB is perpendicular bisector BP. Locus of a point

equidistant form BC is its perpendicular bisector PS. P the point of intersection is the

only point that is equidistant form A, B and C. Hence ‘P’ is the locus.

Example 6: BC is a fixed line AC is a line of variable length and its length varies such

that AC2 = BC

2 + BA

2 BAC and CAB are variable. Show that ABC is a right

angle and the locus of point A(P) is perpendicular to BC.


A1

(P)A

B C

Working: From Pythagoras theorem we have AC2 = AB

2 + BC

2

also given AC2 = AB

2 + BC

2 AC is hypotenuse of the right angled triangle.

AB is BC

Let A another position of the moving point P

Again A1C

2 = A

1B

2 + BC

2

A1BC is right angled triangle AC is the hypotenuse

A1B BC

Hence the locus P is perpendicular to BC

Example 7: ABC is equal to 60o moving point P has a locus such that it is equidistant

from AB and BC line DE is drawn perpendicular to BP at B. Show that at any position of

P ∆DBE is equilateral.

A1

A P1

Q P

60o

B C1

R C

Working: Since P in any position on the locus

BP must be the angular bisector or

o

o

oo

o

o

o

60 BCA

60 BACor

180 BAC2 60

180 BAC2 ABC

)in tringle (Angles --------- 180 BCA BAC ABCBut

CAB BAC and

PC AP

criterion) (SAS --------- PBC ABP

common is BP and

BPC 90 APB

(given) ------------ AC BP Also

PBC ABP ABC


ABC is equilateral triangle

In the new position of ‘P’ also we can prove using the same steps that A ‘BC’ is

equilateral.

Hence any line perpendicular to the locus of P will form an equilateral triangle with line

AB and BC.

Example 8: O is the center of a circle which is the locus of a point P such that

OP = 2cms.Find the locus of a point which is 1cm away from the circle. Find the total

length of the locus of Q.

2 1 Q

O P

Working: Let P be a point on the circle with O as center and radius of 2cms. Extend OP

to Q such that OQ is 2 + 1 = 3cms. If a circle is drawn with O as center and a radius of

3cms. It will represent the locus of Q

Since OQ – OP = 3 -2 = 1cm

Example 9: In the ∆ABC lines AB and AC are extended to D and E. Loci of angles

CBD

and BCE meet at G. Show that line BG is the locus of a point equidistant from AB and

AC.

A

B C

D E

G

Working: BG is locus of a point G that is equidistant from lines BD and BC similarly G

is a point on the locus of a point which is equidistant form BC and CE. Hence G is

equidistant form BD and CE BD and CE are extent ion of lines AB and AC.

Example 10: AB and CD are a pair of parallel line. Show that the locus of a point P

which is the mid point of any transversal across the parallel lines is a line equidistant

form AB and CD.


E G

A B

X p Q Y

C

S

F H

Working: Let EF and GH be any two transversals to AB and CD. Let B and Q be their

mid points. RS is a line through P which is perpendicular to both AB and CD.

Consider triangles PER and PSF

PE = PF - - - - - - - - - - - - - - (given)

EPF = SPE - - - - - - - - - - - - (Vertically opposite angles)

PRE = PSF = 90o

∆PER ∆PSF

PR = PS and RS AB and CD

Hence P is equidistant from AB and CD similarly we can prove Q as equidistant form AB

and CD

Since the distance between AB and CD is constant and PQ is half this distance away form

AB and CD we have AB׀׀ PQ׀׀CD.

EXERCISES

1. Define the term Locus. What is Loci?

2. Select the best answer.

What is locus of a point which is equidistant form three points not in a line?

a) A point

b) A circle

c) A line

d) A pair of lines

3. Find the locus of centers of all circles packing through two given points A and B

show that the locus is the perpendicular bisector of AB

4. What is the loci of the vertices of a right angled triangle given that the vertex

forming the right angle remains unchanged?

5. Find the locus of mid points of equal chords of a circle.

6. ABC is an isosceles triangle given CB ˆˆ , BAD is an exterior angle at A formed

by extending CA. Show that the locus of a point equidistant from BA and AD is

parallel to BC.

7. Find the loci of points P and Q which are equidistant from adjacent points A and

B or B and C given A, B and C are on a line.

8. Q and R are centers of two sets of concentric circles. Show that the locus of the

point of intersection ‘P’ of any two circles with the same radius is the

perpendicular bisector of line AB.


9. P and Q are centers of circle of r1 and r2. Given that circle with center Q just

touches the circle with center P, find the locus of P such that the two circles just

touch each other.

10. AB and CD are two intersecting lines, meeting at O. Prove that loci of points P

and Q which are equidistant form the lines AB and CD, are mutually

perpendicular.

Concurrency

When three or more lines pass through the same point they are said to be concurrent and

the common point is called the point of concurrency of the given lines. There are several

properties of triangles which are associated with concurrency.

Theorem 1: Bisectors of interior angles of a triangle meet at a point.

A

F O E

B D C

Working: consider a triangle ABC let BO and CO be the angle bisectors pr B and C

meeting at point O. Let OF and OE be the perpendiculars to AB and BC also draw OD

BC. If OA can be proved to be the bisector of BAC then all the angle bisectors are

concurrent.

Consider triangles ODB and OBF

OB is common

OBD = OBF ---------- (Construction)

ODB = OFB = 90o

∆ODB ∆OBF

Hence OD = OF Using the same procedure OD = OE .

Therefore O is equidistant form AB and AC. Hence OA dissects angle A. Therefore O is

the point of concurrency of all three angular bisectors. This point is also the center of the

in circle which is a circle with all the three sides as tangent.

Theorem 2: In a triangle the perpendicular bisectors of all the three sides are concurrent.

A

D F

O

B E C


Consider a ∆ ABC in which OD and OF are perpendiculars to AB and AC and O is the

point of intersection of the perpendicular bisectors. The perpendicular to BC from point

O, OE should also bisect OE to prove that all the perpendicular bisectors are concurrent.

Join OB, OA and OC consider triangles OBD and OAD

OD is common DA = DB and

ODB = 90o = ODA

∆OBD = ∆OAD

OB = OA

Similarly we can show OA = OC by considering triangles OAF and OFC

OB = OA = OC

Or OB = OC

Consider triangles OEB and OEC

OE is common

OEB = OEC = 90o

OB = OC ------------ (SAS Criterion)

Hence BE = CE

and BEO = CEO

But BEO + CEO = 180º ---------- (Angles on a line)

BEO =CEO = 90º

Hence OE is the perpendicular bisector of BC

Therefore O is the point of concurrency of the three perpendicular bisectors

Also OB = OA = OC

Therefore we can draw a circle with a radius equal to OA or OB or OC and center O the

circle will pass through all the three vertices. This circle is called the circum circle and O

is the circum center.

Theorem 3: Altitude of a triangle is the perpendicular to a side of a triangle from a vertex

opposite to it. The altitudes of a triangle are concurrent.

D A E

I

O

B C

G

F

H


Construction: Draw a triangle ABC and Altitudes AG, BH and CI. Let O be the point of

intersection of BH and CI. Draw ED׀׀ BC, DF׀׀ AC and EF ׀׀ AD. To prove the

concurrency AG should also pass through O.

Consider triangle ABC and ACE

BAC = ACE --------- (BA׀׀ EC, Alternate angles)

BCA = CAE --------- (BA׀׀ EC, Alternate angles) enough

AC is common Hence, ∆BAC ∆CAE

AE = BC

Consider triangles BAD and BAC

DBA = BAC (AC׀׀ DB Alternate angles)

DAB = ABC (AC׀׀ DB Alternate angles)

AB is common Hence ∆BAD ∆BAC

DA = BC

AE = BC = DA

A is the mid point of DE

Also DE ׀׀ BC Hence AG is also perpendicular to DE

AG is also perpendicular to DE

AG is the perpendicular bisector of the side DE of ∆DEF

Similarly we can prove that BH is the perpendicular bisector of DF and CI is the

perpendicular bisector of CI. Theorem states that perpendicular bisectors of a triangle

sides meet at a point. Therefore the three altitudes AG, BH and CI meet at the point O.

This point is called as the orthocenter of a triangle.

Theorem 4: A line drawn form the vertex of a triangle to the mid point of the side

opposite to it is called the median of the triangle. The medians of a triangle are also

concurrent. A property of the point of intersection is that it divides the median in the ratio

2:1.

Construct a triangle ABC in which the medians BE and CF intersect at a point G. Extend

AG to intersect BC at D and further to point K such that AG = GK .

This theorem can be proved by showing

BD = DC and 1

2

GF

CG

GE

BG

GD

AS

Proof:

A

F G E

B D C

K


Consider triangle AFG and ABK

∆AGK is similar to ∆ABK since all the three angles have the same value.

FG׀׀ GK ---------- (Corresponding angles are equal)

and FG = ½ BK

Similarly we can prove

GE׀׀KC

And GE = ½ KC

Consider triangles BDK and GDC since all the three angles are equal and because BG is

parallel to KC

We have DC

BD

DK

GD

Also because GC is parallel to BK

BC and GK intersect at D

(Detailed proof for this is provided in the unit on parallelograms)

Hence BD = DC AD is a median and all three medians of ∆ABC meet at G

1

2

GF

CG

GE

BG

GD

AG provecan wesimilarly

1

2

GD

AGor 2GD AG or

AG) GK (Given -------------AG 1 GK 2

1But

GK 2

1 DK GD Becuase

Example 1: ∆ABC is an equilateral triangle show that the in center, circum center, the

centroid are all the same point O.

Construction: Draw an equilateral triangle ABC. Mark D, E and F as the mid points of

AB, BC and CD. Let O be the points of intersection of the medians. Now deduce O is

also the in center, circum center and the orthocenter

A

F E

O

B D C

Working: Since D is the mid point of BC

BD = DC

Also ABD = ACD = 60o ----------- (Given ∆ABC is equilateral)


AB = AC -------------- (Given ABC is equilateral)

∆ABD ∆ACD

Hence AD is also the angle bisector of angle A

Similarly BE and CF are also angle

Bisectors or B and C

Hence O is the point of intersection of the angle or the in center of the triangle.

Since ∆ABD ∆ACD

ADB = ADC

But ADB + ADC = 180o ------------ (Angles in a line)

ADB = 90o = ADC

Hence AD is BC

Similarly CF AB and BE AC

Hence O is the point of intersection of perpendiculars from the mid points of sides BC,

AB and AC. There O is the circum center.

The perpendiculars from the vertices of the equilateral triangle ABC bisect the sides

∆ABD ACD

And ADB = ADC = 90o

And BD = DC

Hence O is the point of intersection of the three altitudes or the or two center.

Example 2: Show that in a right angled triangle is orthocenter is concurrent to a vertex.

A

B C

Consider a right angled triangle ABC, with vertex B or the right angle.

Side AB will be an altitude since AB BC

Side BC will be an altitude since BC AB

Let BD be the perpendicular to AC

Hence B becomes the point of intersection of all the three latitudes

Example 3: D, E and F are the mid points of sides AB, BC and CA of ∆ABC. If

D, E and F ate the mid points of sides. AB, BC and CA of a ABC. IF O is the point of

intersection of AD, AE and AF. Show that it is also the point of intersection of the

medians of the ∆DEF. Assume DE׀׀ BC and PE =½ BC, DE =½ AB and DE = ½AB

FD׀׀ AC and FD׀׀ ½ AC

A

F P E

R

B D C

O


Construction: Draw a triangle ABC with D, E and F as mid points of sides AB, BC and

CA. join AD, CF and BE. Let O be the point of intersection. P, Q and R be the points

where DE, EF and DF intersect with the medians.

Given DE ׀׀AB and DE = FA =½ AB

Similarly AE = FD

And FE is common

∆AFE ∆FDE

Now consider triangles AFP and PDF

We have AF = ED -------- (Given)

FAP = PDE ---------- (Alternative angles)

FPA = DPE ---------- (Opposite angles)

∆FAP ∆ PDE

Hence FP = PE

And AP = PD

P is the mid point of FE also lies on AD similarly we can prove Q and R as mid points

of FD and EF. Since B, Q and O, C, R, O are also collinear O must be the point of

intersection of the median of ∆DEF

Example 4: PQR is a triangle with PS, QT and RU as its medians intersecting at the

point O.

a. If QO = 10cm find QT

b. If RO = 7cmx find OU

c. If PS = 15.6cm find QD

P

U O T

Q S R

cms4.106.153

2P

3

2O

3

2

PS

OP c)

cms5.372

1RO

2

1OU

1

2

OU

RO b)

cms15QO2

3QT

3

2

QT

QO a)

:Answer

SP

Example 5: If O is the orthocenter of ∆ABC show that A is the orthocenter of ∆ABC

Construct a ∆ABC and draw perpendicular to the opposite sides from A, B and C. Mark

the point of intersection O.


A

F

O

B C

D

Consider ∆OBC

OD BC - - - - - - - - - - (Given AD BC and O is a point on AD)

OB CE - - - - - - - - - - (Given BE CE and O is a point on BE)

OC BF - - - - - - - - - - (Given CE BF and O is a point on CF)

Hence AO BC

AE OB

AF OC

Therefore A is the point of intersection of perpendiculars opposite sides from the vertices

of triangle OBC. Hence A is the orthocenter of triangle ABC.

EXERCISES

1. Explain with sketches the following terms

a) Orthocenter

b) Circum center

c) Incenter

d) Centroid

2. If the two medians are equal show that the triangle is isosceles and that the

centroid lies on the perpendicular to the remaining side, which is not equal to any

other side.

3. Prove that the angle bisectors of a triangle pass through the same point

(Theorem 1).

4. Perpendicular bisectors of two sides AB and Ac meet at a point O. Show that the

perpendical OD to line BC, bisects the line BC at D.

5. If the medians of a triangle are equal show that the triangle is equilateral.

6. D, E and F are mid points of BC, BA and AC of ABC prove that the orthocenter

of ∆DEF is the circumcenter of ∆ABC.

7. Given sum of two sides of a triangle is always more than the third side show that

sum of any two medians is greater than the third median.

E


QUADRILATIERALS AND PARALLELOGRAMS

QUADRILATERALS

A quadrilateral is a four sided figure. It has four vertices and four interior angles. The

figure below shows a quadrilateral and its features.

A

B

C

D

Quadrilaterals are of different types

A quadrilateral with two sides parallels

is called Trapezium

A quadrilateral in which two

adjacent sides are equal

A quadrilateral in which opposite

sides are parallel is called as a

parallelogram

Sides: AB, BC, CD and AD

Vertices: A, B, C and D

Interior

Angles: D and C,B,A

Diagonals: AC and BD


Theorem 1: Sum of interior angle of a quadrilateral is 360º

Construction: Construct a quadrilateral ABCD join the Diagonal AC and BD

A

B

D C

Proof: consider ∆ADB

D A B + ADB +ABD = 180º

----------- (Angles in a triangle)

consider ∆BCD

A quadrilateral parallelogram

with each of the interior angles

equal to 90º is called a Rectangle

A square a rectangle

with equal sides interior

angles of square is equal

to 90º

A Rhombus is a parallelogram

with opposite sides equal.

Interior angles of Rhombus is

not equal to 90º


BCD + BDC + DBC = 180º

----------- (Angles in a triangle)

Therefore D A B + ADB + ABD =180º

+ (BCD + BDC + DBC) = 180º

DAB + BCD + (ADB +BDC) + (ABD + DBC) = 360º

But ADB +BDC = ADC ---------------- (Adjacent angles)

ABD + DBC = ABC -------------- (Adjacent angles)

DAB + BCD + ABD + DBC = 360º

Hence the interior angles of a quadrilateral add up to 360º

Example 1: Four angles of a quadrilateral are 110º 70º 60º and 120º name the type of

quadrilateral given that the angles are in the order D and C,B,A

Construction: construct a quadrilateral ABCD with A = 110º, B = 70º, C = 60º and D =

120º

Select convenient length for sides AB and BC

A D

B C

A + B = 180º = (110º + 70º)

AD ׀׀ BC ------------- (cointerior angles add up to 180º)

Similarly C + D = 180º

Confirms AD׀׀ BC

However A + D = 110º + 120º = 230º

Hence A + D 180º

AB is not parallel to CD

Therefore two sides of the quadrilateral are parallel and it is a Trapezium.

Example 2: In a kite shaped quadrilateral show that the opposite angles formed by pairs

of unequal sides are equal


B

A C

D

Working: Consider ∆ABC

We have AB = BC

BAC = BCA ------------------- (∆ABC is isosceles)

Consider ∆ACD

We have AD = CD

DAC = DCA ------------------ (∆ACD is isosceles)

Therefore BAC + DAC + BCA + DCA

ButBAC + DAC = BAD ------------ (Adjacent angles)

And BCA + DCA = BCD ----------- (Adjacent angles)

Hence BAD = BCD

Example 3: In a quadrilateral PQRS the opposite angles R and P are equal. If angles Q

= 70º and S = 50º find all the angles of quadrilateral.

Given P + Q + R + S = 360º

P + 70º + R + 50º = 360º

Or P +R = 240º

But P = R ------------ (Given)

2 P = 240º

P = 120º = R

Hence P = 120º, Q = 70º, R = 120º and S = 50º

Example 4: Show that the exterior angles of a quadrilateral also add up to 360º

H

A

B

E

G

D C

Construction: Draw a kite ABCD such

that AB = BC and AD = DC

Join AC

To show BAC = BCD

Construction:

Draw a quadrilateral ABCD

Extend side AB to E, BC to

F, CD to G and GA to H.

To prove: HAB +

CBE + FCD + ADG

= 360º


Working:

HAB + BAD = 180º ------------ (Angles on a straight line)

CBE + ABC = 180º ------------ (Angles on a straight line)

FCD + BCD = 180º ------------ (Angles on a straight line)

ADG + ADC = 180º ------------ (Angles on a straight line)

HAB + CBE + FCD + ADG + BAD + ABC + BCD + ADC +

720º

But BAD + ABC + BCD + ADC = 360º ---------- (Angles in a quadrilateral)

HAB + CBE + FCD + ADG + 360º = 720º

HAB + CBE + FCD + ADG = 360º

Hence exterior angles add up to 360º

Example 5: Any pair of adjacent angles in a quadrilateral add up to 180o. Find all the

four angles of this quadrilateral ABCD given A = 80o

Given A + B = 180o

C + B = 180o

A – C = 0

A = C

A = 80o = C

A + B = 180o

B = 180o – 80

o = 100

o

C + D = 180o

D = 180o -80

o = 100

o

A = 80o, B = 100

o, C = 80

o, D = 100º

EXERCISES

1. Draw examples of following types of quadrilaterals.

a. Trapezium

b. Kite

c. Rectangle

d. Rhombus

2. Three angles of a quadrilateral are 100o, 90

o and 80

o find the remaining angle.

3. adjacent ofpair One .o180 upto )CA( )D B( ralquadrilate a of angles Opposite

.D and C ,B ,A angles theFind .150 and 170 upto add )DC( angles oo

4. Four angles of a quadrilateral measure x o, 2x

o, 3x

o and 4x

o. Find the values of

Show that the diagonal QS bisects angles S and Q .

5. Two angles of a quadrilateral are equal to 70o. Remaining two angles are also

equal. Find their values.

6. Sides AB, BC, CD and DA are extended to points E, F, G and H. If three of the

exterior angles are 100o, 90

o and 80

o find all the interior angles of the

quadrilateral.

7. A Trapezium ABCD shown below is divided to form two quadrilaterals some of

the angles have been marked. Find all the remaining angles.


A E B

1 2 3

100o

5 85o 4 60

o

D F C

8. PQRS is a quadrilateral. Bisectors of angles P and Q meet at T. Show that

PRS + SRP = 2 PTQ

R

S

T

Q

P

9. ABCD is an isosceles Trapezium because D = C and AB ׀׀ CD show that

AD = BC

PARALLELOGRAMS

Theorem 1: In a parallelogram show that opposite sides and angles are equal

A B

C D

Construction: Draw a parallelogram with AB׀׀CD and AC׀׀ BD join

Proof: DCB = ABC ------------ (Given AB׀׀CD)

Also ACB = CBD ------------- (Given AC׀׀BD)

BC is common

ΔACDABC ---------- (ASA criteria)


Hence AB = CD

And AC = BD

Also BAC = BDC

Similarly we can prove ACB = ABD

Converse of this theorem is also true if the opposite sides and angles are equal

quadrilateral is a parallelogram

Theorem 2: Diagonals of parallelogram bisect each other.

A B

O

D C

Construction: Draw a parallelogram ABCD with sides AB׀׀ CD and AD׀׀ BC join the

diagonals AB and CD. Let O be their point of intersection.

Proof: consider triangle OAB and OCD

OAB = OCB ------------- (Alternative angles AB׀׀ CD)

OBA = ODC ------------- (Alternative angles AB׀׀ CD)

AB = CD ------------ (Given ABCD is a parellogram)

∆AOB ∆COD ------------ (ASA criteria)

Hence OA = OC

And OB = OD ------- (corresponding parts in corresponding triangles)

Converse of this theorem is also true. If the diagonals of a quadrilateral bisect each other

then it is parallelogram

If the parallelogram is a rectangle the diagonals bisect each other and the diagonals are

also equal in length.

Theorem 3: In a Rhombus show that the diagonal are perpendicular to each other.

A

O

B C

D


Construction: construct a Rhombus such that AB = BD = DC = CA. Join the diagonals

AD and BC Let them intersect at point O. Prove AO BC

Consider the two triangles AOB and AOC

AB = AC ---------------- (Given ABCD is a Rhombus)

BO = OC ---------------- (Because ABCD is a parallelogram its diagonals bisect each

other)

AO is common

AOCAOB

ΔAOCΔAOB

But AOB + AOC = 180o

2AOB = 180o

orAOB = 90o = AOC

Hence AO BC

The converse of this theorem is also true therefore if the diagonals of a quadrilateral

bisect at right angles, it will be a Rhombus.

In the case of a square which can be considered as a Rhombus with equal sides, the

diagonals are equal and bisect each other at right angles.

Example 1: State any four important properties of a parallelogram

Answer:

a. Opposite sides of a parallelogram are equal

b. Opposite angles of a parallelogram are equal

c. Diagonals of a parallelogram bisect each other

d. Any two adjacent angles of a parallelogram add upto 180o

Example 2: Given one of the angles A = 70º in a parallelogram ABCD find the remaining

three angles.

Working: Adjacent angle B = 180º - A

------- (AB ׀׀ CD and BC is a transversal)

B =180º -70º = 110º

AC = 70º --------- ABCD is a parallelogram and C is opposite to A )

Also BD = 110º ------------- (ABCD is a parallelogram and C is opposite to A )

Example 3: Shown below is rectangle ABCD with diagonals AC and BD intersecting at

O if CO measures 3cm find the length of diagonal BD

A B

3cm

O

D C


Working: Consider triangle ABCD and ACD

We have AD = BC ------------------- (ABCD is a rectangle and AD and BC are opposite

sides)

DC is common

ADC =BCD = 90º ------------- (ABCD is a triangle)

∆AOC ∆BCD

Hence AC = BD

Also AC and BD bisect at O

Hence AC = BD = 2CO

Or AC = BD 2×3 = 6cms

Example 4: PQRS is a Rhombus if PQR = 60º show that ∆PQS is equilateral

Q

P R

S

Consider ∆PQR

PQR = 180º - QPR – QRP ------------- (Angles is a triangle)

But QPR =QRP ------------- (because QP = QR)

PQR = 180º – 2QRP

2QRP = 180º – PQR = 180º- 60º =120º

QRP = 60º = QPR

PQR is an equilateral triangle since all the three interior angles are equal to 60o

Example 5: ABCD is a parallelogram E and F are the mid points of parallel sides AB and

CD. Show that EC׀׀ AF and AECF is a parallelogram.

A E B

D F C

Construction: Draw a parallelogram ABCD with AB׀׀ CD and AD׀׀ BC. Mark mid

points of AB and CD as E and F. Join AF and EC


Working: Consider triangles ADF and BEC

We have ADF =EDB ---------- (opposite angles of a parallelogram)

AD = BC --------------- (opposite sides of a parallelogram)

DF = 2

1DC = AE

2

1= (AB) ------------ (AB and DC opposite sides of a parallelogram)

∆ADF ∆ADB --------- (SAS criteria)

AF = EC

And AC = FC -------- (E and F are mid points of AB and DC)

AEFC is a parallelogram since opposite sides are equal

Example 6: Two angles of a parallelogram are x + 30º and x - 30º. Find the value of x

and all the angles of a parallelogram.

A B

x + 30º

x – 30º

D C

Construction: Draw a parallelogram ABCD. Let A = x + 30º and D = x – 30º

Working: Since AB ׀׀ DC

We have A + D = 180º ------------- (cointerior angles)

x + 30º + x -30º = 180º

2x = 180º

x = 90º

C = A = x + 30º = 60º + 30º

C = 90º

B = D = x – 30º = 60º – 30º

B = 30º

A = x + 30º = 90º + 30º = 120º

D = x – 30º = 90º – 30º = 60º

Example 7:

ABCD is a square AC and BD are two diagonals intersecting at point O. Prove that

OAB = OBA = 45º

A B

O

D C


Construction: Draw a square ABCD join diagonals AC and BD let O be their point of

intersection.

Working: Consider triangle AOD and AOB

AD = AB -------------- (sides of a square)

DO = OB ------------- (ABCD is also a parallelogram and diagonals bisect each other)

AO is common

∆AOD ∆DOC (similarly ∆AOB ∆AOD ∆ODC ∆OBC)

AOD = AOB

But AOD +AOB = 180º

Or 2 AOB = 180º

AOB = AOD = 90º

Consider ∆AOB

OAB + OBA + AOB = 180º ---------- (Angles in a triangle)

OAB + OBA = 90º ------------ (AOB = 90º)

But OA = OB -------------- (∆AOB ∆AOD ∆AOC)

AOB is isosceles

Hence OAB =OBA

OAB + OBA = 90º

Or 2 OAB = 90º

OAB = 45º

Hence OAB = OBA = 45º

Example 8: ABCD is a rectangle OA and OB are angle bisectors or A and B. Show that

AOB = 90º

A B

O

D C

Working: DAB = CBA = 90º --------------- (ABCD is a rectangle)

OAB =2

1 DAB =

2

1 CBA = OBA --------------- (OA and OB are bisectors)

OAB + OBA = DAB = 90º

But OAB OBA + AOB = 180º

AOB =180º – 90º = 90º

Example 9: Following figure shows a parallelogram ABCD with a diagonal AC. Find all

the angles of the parallelogram given CAB = 40º and CAD = 80º


D C

80º

40º

A B

Given CAB = 40º and CAD = 80º

We have CAB + CAD = 40º + 80º = 120º

DAC = 120º

ADC + DAC = 180º ------------ (cointerior angles AD׀׀ BC)

ADC + 120º = 180º

ADC = 60º

DCB = 120º = DAC ----------- (Opposite angles in a parallelogram)

And ADC = 60º = ABC --------- (Opposite angles in a parallelogram)

Example 10: ABCD is parallelogram. EF is a line joining mid points of sides AD and

BC. GH is a line joining any two points on lines AB and CD. O is the point of

intersection of EF and GH show that DG = OH

Construction: Draw a parallelogram ABCD mark E and F as mid points of AD and BC.

Mark any point G on AB and H on DC join EF and GH. Mark O as the point of

intersection. Draw IJ ׀׀AD and BC passing through O.

A I G B

O

E F

D H J C

Working: Consider Triangles ABE and BEF

2

1AD = AE = BF =

2

1BC ---------- (Given E and F are mid points)

BE is common

∆ABE ∆EBF

Hence AD = EF

Also AB׀׀ EF׀׀ DC

Consider the quadrilateral AIOE


AE׀׀IO -------- (Given)

AI EO ---------------- (AB׀׀ EF and I and O are points on AB and EF)

AIOE is a parallelogram

AE = IO = 2

1AD =

2

1 BC

Similarly we can show

ED = OJ = 2

1AD =

2

1BC

IO = OJ

Consider triangle IOG and HOJ

IO = OJ

OIJ = OJH ---------- (opposite angles)

OHJ = OGI ----------- (Alternate angles)

∆OHJ ∆OGI

GO = HO Hence EF bisects GH

EXERCISES


a. In a parallelogram opposite sides are equal

b. In a parallelogram opposite angles add upto 180º

c. Diagonals of a Rectangle bisect the vertex angles

d. Interior angles of a square is equal to 90º

e. Sum of exterior angles in a parallelogram is 360º

f. Interior angles of a Rhombus are all equal

g. A parallelogram is a quadrilateral

h. Trapezium is a type of parallelogram

i. Diagonals of a Rhombus are mutually perpendicular

2. Interior anglesA of parallelogram is 55º side AB measures 3cms the perimeter

of ABCD is 16cms. Find all the angles and length of all the sides.

3. PQRS is a Rhombus length of the diagonal PR is equal to the length of a side.

Show that Pr divides the Rhombus into two equilateral triangles.

4. ABCD is a rectangle. Angle bisectors of A and B meet at O. Show that AO

is Perpendicular to BO.

5. ABCD is a parallelogram and side AB is equal to diagonal AC. If side DA is

extended to E show that AB bisects EAC.

6.

E

A B

D F C


In the figure shown above ED and FB bisect angles B and D of a parallelogram

s show that DE BF and DE = BF.

7. Show that a parallelogram is a square if its diagonals are equal and intersect at

right angles.

8. E is the mid point of side AB of a parallelogram ABCD. If CE bisects BCD

show that DE also bisects ADC. Find the value DEC.

9. ABCD is a parallelogram side DA is extended to E such that DA = EA and EC is

joined. Show that AF =2

1 DC =

2

1AB and F is the mid point of EC.

E

A B

D C

10. E and F are mid points of AB and CD of parallelogram. AF and CE, and DB

(diagonal) Joined AF intersects DB at x and CE intersects DB at y. Show that Dx

= xy = yB =3

1 DB


AREAS OF PLANE AND SURFACE AREA OF SOLID FIGURES

AREA OF PLANE FIGURES

Consider a plane figure such a polygon or a closed curve. Area is a measure of the size

interior of this figure. The figure 12.1 shows a Rectangle made from flat square tiles.

Simplest way to express the area is to total the number of square tiles involved in making

this rectangle. If the square tiles have sides equal in length to a unit of measurement for

e.g. cm or mm, the total number of squares will be the area of the figure in mm mm or cm

cm (mm2 or cm

2)

1 2 3 4

5

10 9 8 7 6

11 12 13 14 15

Figure above shows a rectangle made from 15 square tiles and therefore can be side to be

of 15sq units size. Since the layout is 5 tiles long and 3 tiles broad area can also be

expressed as lb or 5×3 = 15sq units

Irregular shapes for e.g. Scalene triangle may also need a number of fraction or part tiles

to build. The whole number of tiles and the sum of part tiles can be added to obtain the

area of this figure. Figures can be graph sheet and squares counted. However formulas

can be to calculate the areas of polygons, circles and similar shapes.

Every polygon has a part of the plane enclosed by it. This part with the polygon is

defined polygon region. Every polygon region has an area which is positive and real and

is measured in square units. However, the sides of the polygon has only length is its

dimension and has no width.

If two polygons are congruent, their areas are also equal.

Theorem 1: A parallelogram can be divided into two triangles having the same area by

its diagonal.

A B

Rectangle from 15 square tiles

D C


Construction: Draw a parallelogram ABCD and join BC.

Proof: Consider triangles ABD and BCD

AB = CD - - - - - - - - - - (ABCD is a parallelogram)

AD = BC - - - - - - - - - - (ABCD is a parallelogram)

BD is common ∆ABD ∆BCD - - - - - - - - - - (SSS criterion)

Hence Area of ∆ABD = Area of ∆BCD

BC divides a parallelogram into two parts.

Theorem 2: Parallelogram with a common side and between the same parallel lines have

the same area.

E C F D

A B

Construction: Draw a parallelogram ABCD. Use AB as the common side or the base

and draw another parallelogram ABFE. Points E and F should be on the line AB or

extention of AB.

Proof: Consider triangles AED and BCF

AE = BF - - - - - - - - - - - - (Given ABEF is a parallelogram)

AD = BC - - - - - - - - - - - - (Given ABCD is a parallelogram)

ADE = BCF - - - - - - - - - - (Corresponding angles AD BC)

AED = BFC - - - - - - - - - - (Corresponding angles AE BE)

∆AED ∆BCF - - - - - - - - - - - (ASA criterion)

Hence area of ∆ADE = area of ∆BCF

Consider parallelogram ABCD

Area of ABCD = Area of Trapezium ABDF + Area of ∆BCF similarly

Area of ABEF = Area of Trapezium ABDF + Area of ∆AED

Given area of ∆AED = Area of ∆BCF and that ABCD is common to both the

parallelograms

Area of ABEF = Area of ABCD

Theorem 3: Triangle with their base as a common side and with their remaining vertices

on a line parallel to the base have the same area.

E A D F

B C


Construction: Draw two triangles ABC and DBC such that AD is parallel to BC. Extend

DA to E such that BE׀׀ AC and AD to F such that BD׀׀CF

Proof : Consider parallelograms ACBE and DBCF

Area of ACDE = Area of DBCF

Given that two parallelograms have a common base BC and the sides opposite to BC are

on the same line EF.

Area of ACBE = Area of ∆ACB + ∆ABE

But area of ∆ACB = area of ∆ABE

Given ACBE is a parallelogram and AB is a diagonal

Area of ACBE = 2 ×area of ∆ABE

Similarly Area of DBEF = 2× area of ∆BCD

2× area of ∆ABE = 2× area of ∆BCD

area of ∆ABE = area of ∆BCD

AREA FORMULAE – 1. RECTANGLE

AXIOM: Area of a rectangle is defined as l ×b units. Where l is the length (measure of

longest side) and b is the breadth (measure of the shortest side)

PARALLELOGRAM

Area of a parallelogram can be calculated by comparing its area to that of a rectangle.

Consider the following figure. ABCD is a parallelogram. A rectangle is drawn with one

of its sides common to one of the sides of the parallelogram. Base of the parallelogram

AB can be selected as the common side perpendiculars from A and B are drawn and

points F of intersection of these perpendiculars with line CD or extention of it are marked

as E and F. ABEF is the rectangle contained between the same parallel lines of the

parallelogram.

D E C F

A B

Area of parallelogram ABCD = Area of rectangle AEFB

Area of parallelogram ABCD = AE×AB - - - - - - - - - - ( l ×b AEFB is a rectangle)

AB = Base of the parallelogram and

AE = height of distance between the opposite sides

Area of parallelogram = base× height = b× h


TRIANGLE

Area of a triangle is half the area of a parallelogram formed by two sides and their

parallel lines with the remaining side forming a diagonal.

A D

B E C F

Consider a ∆ABC drawn as a part of the parallelogram ABCD.

bh2

1

base height2

1

EF AE2

1 ABC of Area

rams)parallelog of (Opposite ------- BC AD EF Also

BC) (AE ------- triangleofheight AE

EF AE

ADEF Rectangle of Area ABCD of AreaBut

ABCD) of diagonal is (AC ------

ABCD ramparallelog of Area2

1 ABC of Area

Example 1: Show that diagonal of a parallelogram divides it to four triangles of equal

area.

A B

F

O

E

D C


Construction: Draw a parallelogram ABCD. Join diagonals AC and BD.

(area) AOB AOD of Area

AE OB2

1 AE OD

2

1

ABCD) of diagonal theis (BD --------- OB ODBut

AE OB 2

1height base

2

1 AOB of Area

AE OD2

1height base

2

1 AOD of Area

AOB and AEO rianglesConsider t :Answer

O.at intersect BD and AC BD, CF and BD AE Draw

Similarly we can prove that

Area of AOB = Area of ∆BOC = Area of ∆COD

Example 2: Show that the area of a Rhombus is equal to the product of 2

1lengths of two

diagonals.

B

O

A C

D

Construction: Draw a Rhombus ABCD join AC and BD. O be the point of intersection.

AC BD 2

1

AC DO) (BO 2

1

AC DO 2

1AC BO

2

1 ABCD of Area

(Area) ADC2

1 (area) ABC

2

1 ABCD Rhombus of Area

base hight 2

1 AC DO

2

1 ADC of Area

baseheight 2

1 AC BO

2

1 ABC of Area :Answer


Example 3: ABCD is a quadrilateral diagonal AC measure 5cms. BE and DF are

perpendiculars to AC. Given BE = 3cms and DE = 2cms. Find the area of the

quadrilateral.

B

A F

E C

D

Construction: Draw a quadrilateral. Draw AC and BE and FD (Quadrilateral will not be

to any scale)

2

2

2

12.5cm 5 7.5 ABCD of Area

ACD of Area ABC of Area ABCD ralquadrilate of areaBut

5cm252

1

FDAC2

1heightBase

2

1 ACD of Area

7.5cm352

1 ABC of Area

BE AC2

1 height base

2

1 ABC of Area

ABCConsider :Answer

Example 4: ABCD is a parallelogram. E and F are mid points f AB and CD. Show that

area of AEFD is half of area of ABCD.

Construction: Draw a parallelogram ABCD. Mark E and F as mid points of AB and CD

join EF Draw EG CD.

A E B

D F G C

Working: Consider the quadrilateral AEFD


AE׀׀ FD - - - - - - - - (Given AB׀׀CD)

Example 5: ABCD is a trapezium. Prove that area of a trapezium is its height times the

average length of the two parallel lines.

Construction: Draw a trapezium ABCD. Draw perpendiculars form A and B to CD.

Let AE and BF be the perpendiculars.

A B

C E F D

ABFE Rectangle of Area and BDF of Area ACE of Area trapeziumof Area

EF AE AE FD 2

1 AE CE

2

1

ABCD trapeziumof Area

EF AE ABFE Rectangle of Area

BF) -(AE--------- AE FD 2

1 BF FD

2

1 BFD of Area

AE CE 2

1 ACE of Area

ABCD ramparallelog of Area2

1

EG CD 2

1

height) (base -------EG DF

AEFD ramparallelog of Area

height) (base --------EG CD

ABCD ramparallelog of Area

parallel and equal are sides opposite since ramparallelog a is AEFD

CD AB

ram)parallelog a is ABCD(Given ------- DF CD2

1 AB

2

1 AE


2

CD AB AE Trapezium of Area

)(-----------2

FD

2

CE

2

2EF

2

FDCEADAB

2

CDABBecasue

2

CDABEFFD

2

1CE

2

1But

EFFD2

1CE

2

1AE

EFAB

EXERCISES

1. Find the area of the following figures

a. A triangle with base 10cms and height 5cms.

b. A parallelogram with base 10cms and height 7cms.

c. T trapezium with its parallel sides equal to 5 and 7cms and height 8cms.

d. A right angled triangle with sides 3cm, 4cm and 5cms.

e. A Rhombus with diagonals measuring 12cms and 15cms.

2. Show that a median divides a triangle into two parts having the same area

3. Following figure shows a parallelogram ABCD, and AECD and CFAD.

Given AE = 5cms, CD = 7cms and CF = 6cms Find AD.

A B

5cms

6cms

D 7cms C

4. ABCD is a quadrilateral and AC is a diagonal. If O is mid point of AS show that

area of quadrilateral AOCD = area of equilateral AODB.

5. PQRS is a parallelogram. T and U are mid points of PQ and RS. RT, TU and RS

are joined to divide the parallelogram into four triangles given ∆PST has an area

of 5sq cms find the areas of remaining triangles and the parallelogram.

S U R

P T Q


6. P, Q and R are the mid points of an equilateral triangle ABC. Show that the area

of ∆PQR is quarter the area of ∆ABC.

7. To isosceles triangles ABC and BCD are drawn with base BC as common and A

and D are equidistant from BC and on opposite sides of it. F in the area of

∆ABD, ∆ACD and Rhombus ABDC given BC = 2cms and AD = 3cms.

8. ABCD and ABEF are two parallelograms having AB as common side and C, D, E

and F are in the some line. FB is joined given AB = 12cms and ∆FAB = 120cm2

find the following:

D F C E

A B

a) Area of both the parallelogram

b) Height of the parallelogram

c) Length of side AB

9. ABCD is a rectangle DC is extend to E such that ABCE is a parallelogram. If the

dimensions of the rectangle are 8cms = AB and CD = 6cms find

a) Area of the parallelogram

b) Areas of ∆ABC, ∆ADC and ∆CBE

D C E

A B

10. ABCDEF is a regular Hexagon. Two of its diagonal AD and BF measures 4 and

2 3 cms. Find the area of the Hexagon and length each side given all the sides

are equal.

B C

A B

F E


PERIMETER AND AREA

Perimeter of a plane figures like polygons is sum of the length of all the sides. In the case

of closed curves it is the length of the bounding perimeter is expressed in the same units

of length for e.g. cm or mm. Area of polygon and closed curves like circles can be found

if the perimeter and some of their properties are known.

Consider a triangle ABC a, b and c denote the sides BC, AC and AB of the triangle

The perimeter P = a + b + c

Semi perimeter of half the value of perimeter

S = ½(a + b + c)

Heros formula: A formula named after ‘Hero’ connects the dimension pf the sides of a

triangle to its area.

))()(( csbsssAreaA

Quadrilateral: Area of a quadrilateral can be found from the length of all the four sides

and one of its diagonals.

B

A

C

D

Diagonal AC divides the quadrilateral into two triangles. Area of each triangle can be

found using Hero’s formula. Adding the areas of the triangle gives the area of the

quadrilateral special quadrilaterals for e.g. Rectangles area and perimeter can be linked

by other formulae also.

equal) are sides all of (lengths ------ 4 Perimeter

diagonals oflength are d and d where,d d Area

Rhombus

4 Perimeter

equal) are sides all of (lengths -------- Area

Square

equal) are sides (opposite -------- 2b2Perimeter

)definition(by -------- b Area

Rectangle

2121

2

l

l

lll

l

l

Circle

Perimeter of a circle is known as circumference by definition of A

The circumference 2πD π circle a of C'' r - - - - - (D = 2r)


Where D is the diameter or the longest chord (line) intersecting the circle at points and

passing through its center.

radius)2(D ------πr4

πD circle a of Area 2

2

Part of the circumference is called as the are of a circle. When the ends of an arc are

joined with center of the circle by two radii sector of the circle is formed.

Perimeter of an arc is given as

radii theseperating angle theis where360

2πθrS

o

o

Since, if we consider the entire circumference which has a perimeter of 2π r as the arch

the radii are separated by 360o

o360

2r2r P

sector a ofperimeter Hence

r

o

2

360πrsector a of Area

Segment of a circle is the part of a circle bound by a line intersecting the circle at two

points. The line divides a circle in to major and minor segments. This line is called as a

chord. When the circle is exactly divided into two parts this line or chord will be

diameter. Half of a circle is called a semi circle.

O

A C B

OAB of Area - OABsector of Area :segment of Area

chord theoflength theis C and OB andOA radii by two

formed angle theis whereC,360

segment a ofPerimeter

o

r


Example 1: Figure below shows shaded cross B and of a national flag. Find the area of

shaded and unshaded part. Find the perimeter of the flag.

10-50cms

2

1 10cm 30cm

3

10cm

Example 2: ABC is an equilateral triangle, side AB is 2cms. Find the perimeter and the

area of the triangle. From the area find the height of the triangle.

A

B D C

Construction: Draw a triangle ABC with all the sides equal to 2cms. Draw AD BC

Perimeter of ∆ABC = AB + BC + CA

= 2 + 2 + 2 = 6cms

Semi perimeter S = 3cm2

6

Area of the triangle as per Hero’s formula

160cms

30 2 50 2

2 2 flag theofParimeter

800cm 700 - 1500 part unshaded of Area

(3) and (2) (1), regions Adding

700cm 10 10 10 10 50 10

repition any excludingregion shaded of Area

1500cm 3050 flag of area Total

2

2

2

)(

bl


1.73cm Height

Height 22

1 1.73

Height base 2

1 A But

1.73cm 3 A

2)-(3 2)-(3 2)-3(3 A

2cms AB side c

2cms AC side b

2cms BC Side a

3cms perimeter Semi s where

c)b)(sa)(ss(sA

Example 3: PQRS is a parallelogram with PQ and QR measuring 34 and 20cms. The

diagonal QRS measures 42cms. Find the perimeter and the area of the parallelogram.

P 34cms Q

42cms 20cms

S R

Perimeter = 2 (20 + 34) - - - - - - - - (PQRS is a parallelogram)

= 108cms

34cms PQ c

20cms PSb

42cms QS a

48cms cms2

97

2

42 34 20

PQR ofperimeter semi s where

c)-(s b)-(s a)-(s s PQS of Area

PQR of Area 2 PQRS of Area


2336cm

146 4

14 14 36 16

14 14 2 6 6 8

14 28 6 48

34)-(4820)-(4842)-(48 48 PQS of Area

Example 4: ABCD is a rectangle sheet of size 30cms 10cms. A semicircle of Diameter

10cms has been cutout from both the smaller edges. Find the remaining area of the sheet.

Also find the perimeter.

Construction: Draw a rectangle (not to scale) with sides in the ratio of 3:1. Mark the

center of the shorter sides BC and DA. Draw semi circles.

A B

D C

10cm) r hence 20cms (BC --------- 2

10

7

22

2

r semicircle of Area

cm600

2030

ABCD rectangleuncut of Area

2

2

2

BCABbl

2

2

290cm

155 2- 600

cuttingofter ABCD of Area

155cm 50 7

22

Example 5: Find the area of the shaded part of the sector given inside radius = 15cms,

outside radius = 2.5cms and the angle of separation of the two radii = 30o. Also find its

perimeter.


Construction: Draw a sector of a circle with out radius of 2.5cms and 30o as angle of

separation. Draw the inner arc with a radios of 15cm shade the portion enclosed by the

two arcs and the radii.

D

O

C B

2

2

2

2

2

o

2

1.05cm

0.59-1.64 ABCD of Area

0.59cm

360

30 1.5

7

22 ODCsector of Area

1.64cm

60

30 2.5

7

22

360

θπr AOBsector of Area

4.1cm

0.79 1.31 ABCD ofPerimeter

1cm 1.5- 2.5 CD BC

0.79cm

360

301.5 2 CD Arc

cm31.1

360

30 2.5 2 AB Arc

DA CD Arc BC AB Arc ABCD ofPerimeter

EXERCISES

1. Area of a rectangle is 9.8cm2. Given its length equals twice breadth, find the

dimentions of the rectangle.

2. Sides of a triangle are in the ratio x:2x:3x. If the perimeter of the is 36cms find its

dimentions and area. Find the shortest height or altitude of this triangle.

A


3. A right angle triangle has a perimeter of 24cms and an area of 24cm2. Given the

length of an adjacent side of the right angle as 6cms. Find the lengths of

remaining two sides.

4. Diagonal of a quadrilateral measures 8cms. Perpendiculars drawn from the

vertices opposite to the diagonal measures 6cms and 4cms. Find the area of the

quadrilateral.

5. Find the perimeter and area of the following figure.

35

10

30

10

45 1-15

6. The flag shown below comprises of the shaded circle of 20cms diameter at the

center of 30cms by 90cms rectangle. Find the areas of shaded and unshaded part.

Also find the perimeter of the flag.

7. Circumference of a circle exceeds the diameter by 16.8cms. Find the

circumference and the area of the circle.

8. Perimeter of a semicircle is 90cms. Find its radius and the area.

9. A cyclist goes around a track of 100m diameter. Given the diameter of the cycle

wheel as 1m and that the cycle travels a distance of or meters per revolution find

the number of revolution of the cycle wheel to go round the track once. Also find

the length of the track.

10. Following figure shows two circles of 6cms diameter intersecting at two points.

Given that the centers of the circles are 4cms apart. Find the shaded common area.


SURFACE AREA AND VOLUMES OF SOLIDS

Exterior of a solid comprises of a number of plane or curved surfaces. Area such

surfaces can be found using formulae in the case of surfaces having a regular Geometric

shape for example prisms and pyramids.

A prism has two identical polygons as base and top joined by edges forming rectangular

lateral surfaces.

A pyramid has a polygon as its base. Sides of polygon form the base of triangles.

Vertices of these triangles meet at a common point which becomes the vertex of the

pyramid. The side faces or the triangles become the lateral surfaces.

Volume of a solid is the total space with in the surfaces.

1. Write the formulae for finding L.S.A, T.S.A and volume of Pyramids and Prisms.

2. Find the lateral and total surface area of a prism whose height is 4cm and

perimeter of the equilateral triangle base is 6cm.

3. Height of a right angle triangle base and top prism is 15cms. Its base has

dimentions of 13cm, 14cm and 15cm. find its lateral surface area, total surface

area and volume.

(Hint: Use Quadratic equation)

4. Total surface area of a square based prism is 1000cm2. Given its height as 20cm.

find the dimentions and volume of the prism.

5. Base and top of a prism are congruent isosceles triangles of area 192cm2,

perimeter 64cm and the side which forms the base of the triangle 24cms. If the

height if 30cms find its L.S.A, T.S.A and volume.

6. A tetrahedron (triangle based pyramid) consists of four equilateral triangles of

6cms each side. Find its L.S.A and T.S.A

(Hint: Add area of triangles)

TRIANGULAR

PRISM

TRIANGUALR

PYRAMID


7. A square based pyramid has an on area of 576 units. If it is vertical height is

16cms. Find it L.S.A, T.S.A and Volume.

8. A fish tank side walls are made from 2 sq 10cms square glass panels and two 10

2cms rectangular glass panels. Top and bottom are made from steel plates. Find

the area of glass sheets and steel plats required. Find the volume of water the tank

ban hold in liters.

9. A tetrahedron is formed by 4 equilateral triangles of sides 2a. Given median of the

equilateral triangle is 3a find the L.S.A and T.S.A of the pyramid.

10. A sky scraper is modeled on a pyramid at the top of a square based prism. Area of

base if 576m2. If the total height is 180m to the top most point, find the L.S.A,

T.S.A and the volume of the sky scraper.


GRAPHS AND GEOMETRIC CONSTRUCTION

Graphs:

Scientific data is represented as graphs show the relationship between two groups. First

group of data where the variable is controlled is represented on the x – axis. The second

group of data which is dependent on the y –axis. A point on the graph sheet (x –y plane)

represent a set of values of both the variable measured as the co ordinates of the point.

For example consider a water tank being filled by water from a tap.

A set of x – values gives the time elapsed after opening the tap and y the amount of

water filled.

X 1 2 3 4 5 6

Y 1.5 3 4.5 6 7.5 9

Y

O X

Example 1: A cup containing boiling water at 100oC is cool till the water is close to room

temperature. Readings of temperature against time is given below. Draw a graph of

temperature vs. time for the first 10 minutes of cooling.

X-

time 0 1 2 4 6 8 10

Y-

temp 100

o 90

o 82

o 67

o 55

o 45

o 36

o

Note: A graph should have a

title x axis and y-axis marked

and indented x –y axis to be

labeled

Volume

in liters

Time in minutes

Time vs. volume of water

1 2 3 4 5 6

7

8

9 10

1

2

3

4

5

6

6

7


Working:

(1) Draw x and y axis of 10cms each.

(2) Indent x axis (scale markings) 1cm = 1minute.

(3) Indent y –axis (scale markings) 1cm = 10oc

(4) Label x and y axis

(5) Mark each point

Point 1 Mark 100oc an the y –axis (x = 0)

2 Draw a horizontal line from 82o on the y –axis and a vertical line from t minutes.

Mark the point of intersection and label the co ordinates 3 -10. Mark all the remaining

points in the same way.

(6) Join the points and title the graph

(0.100)

(2.82)

Example 2: A cyclist goes on a straight road. He accelerates from 0 – 10km/hr in 10

minutes time. He then slows down at an uniform rate and halts after an additional period

of 10 minutes D. Draw the velocity time graph

Y (10.10) (20.10)

10 A B

Km/ph

C (30.10)

0

10 20 30 X

Minutes

Temp in oC

Time in minutes

Temp vs. time

2 4 6 8 10

20

40

60

80

100


Construction:

Explain the graph line OA shows the accelerating part of the cyclists journey. Velocity

of the bike can be obtaining by drawing a vertical line from a point an the x- axis

corresponding the time. Mark the point of intersection of the graph and the vertical line.

Y – coordinate of this point gives the velocity of bike.

Horizontal line AB is the graph of velocity vs. time during the next 10 minutes of the

journey when the velocity was constant.

Line BC shows the decelerating or slowing down to rest part of the journey.

GEOMETRIC CONSTRUCTION

Geometric shapes can be drawn on a sheet of paper given dimentions or sizes of parts of

a line, triangle or a polygon. An instrument box (Geometry box) comprising of a

compass, divider, scale, protractor and set squares is the required tool kit. Unknown

dimensions of a shape can be measured from an accurate construction of that shape.

Example 3: Construct a ∆ABC with sides equal to 6cms, 7cms and 8cms.

Steps:

1. Draw a horizontal line. Mark point B. Mark point C using a compass or a rules

such that the line segment BC = 7cms.

2. With B as the center draw an arc with a radius of 6cms.

3. With C as the center draw an arc with a radius of 8cms.

4. Mark the point of intersection of two arcs as A. Join AB and AC.

A

6cms 8cms

B C

7cms

Draw x-y axis

Mark 10, 20 and 30 minutes on the x-axis at

equal intervals

Mark 5 and 10km/ph at equal intervals on the

y-axis

Mark point A coordinates (10, 10).Point B

(20.10) and point C (30.0)


Example 4: Draw a right angled triangle PQR given hypotenuse QR = 5cms, side PR =

3cms and P = 90o

P

Q R

Construction:

1. Draw a horizontal line segment PR = 3cms.

2. Draw a line perpendicular to PR at the point P

(a) Draw semi circle of point P

(b) Using the same radius draw two consecutive arcs with the point of intersection

of the semi circle and line PR as the center. Mark the points of intersection of

the semi circle with the arcs. Use these points as centers and draw a pair of

intersecting arcs.

(c) Join the point of intersection of the arcs with Q and extend the same.

3. With R as center draw an arc, to intersect the perpendicular to QR, having Q radius of

5cms.

4. Mark this point as P and join PQ and PR

Example 5: Construct a parallelogram with adjacent sides equal to 4cms and 6cms and

an included angle of 60o

Construction:

(1) Draw a line DC = 6cms

(2) Draw an arc with D as center and intersecting DC

(3) Use the above point of intersection as center and with the radius unchanged draw

another arc

(4) Mark the point of intersection of the two arcs. Join D with this point and extend it

to a length of 4cms. Mark the point as A.

(5) Draw an arc of radius 6cms with A as the center.

(6) Use C as center and draw an arc with a radius of 4cms.

(7) Mark the point of intersection of the two arcs as B. Join AB, BC, CD and DA

A B

4

D 6 C


EXERCISES

1. Construct the graph of area of a square against length of its side from the table

given below.

X 0 0.5 1 1.5 2 2.5 3

y 0 0.25 1 2.25 4 6.25 9

2. A car accelerates from 0-30kmph on a straight road in 60secs. It continues to

move for a further period of 2minutes but its speed comes down 25kmph due to

friction. The driver then applied the brakes and brings the car to a halt in

45seconds. Draw the speed time graph.

3. Construct a triangle with three sides of lengths 8, 9 and 10cms.

4. Draw a right angled triangle given the hypotenuse = 10cms and the shortest side =

6cms.

5. Draw a triangle given its base = 5cms and two of the angles adjacent to the base

= 45o and 60

o .

6. Divide a line segment AB (10cms) into 3 (ratio of parts) 2:3:4. (Hint- Draw a line

AC and use a ruler to mark 2, 2 + 3 and 2 + 3 + 4cms. Join C to B and draw

parallel lines to it to intersect AB at D and E. Now AD: DE: EB = 2:3:4)

7. Draw a triangle ABC given BC = 5cms AB + AC = 8cms and ABC = 60o (Hint:

draw two lines DB and BC = equal to 8 and 5cms and DBC = 60o Join DC and

draw a perpendicular bisector to CD to intersect DB at A.

8. Construct a triangle ABC with BC = 10cms, B = 60o, C = 45

o. Draw a circum

circle passing through points A, B and C. (Hint: Draw ABC, Draw perpendicular

bisectors to AB and BC. Point of intersection of the bisectors is the circum

center).

9. Construct a Rhombus of side 4cms and an include angle of 45o.

10. Triangle ABC has perimeter of 14cms and B = 80o and C = 60

o. Draw

ABC. (Hint: Draw a line PQ = 14cms. Draw P = 40o and Q = 30

o, two lines

intersecting at point A. Draw perpendicular bisectors to PA and QA intersecting

PQ at B and C. Join AB. BC and CA).

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Date post:	13-Mar-2018
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Ganit Learning Guides Intermediate Geometry lines in a plane can have more than one point in common....

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