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# SUBMITTED BY: Name: MUHAMMAD RAFIQUE SKB No.: 136 Department: Mathematics - PowerPoint PPT Presentation

Presentation. PRACTICAL GEOMETRY. SUBMITTED BY: Name: MUHAMMAD RAFIQUE SKB No.: 136 Department: Mathematics Section : Secondary. SHEIKH KHALIFA BIN ZAID COLLEGE. PRACTICAL GEOMETRY.

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PRACTICAL GEOMETRY

SUBMITTED BY:

SKB No.: 136

Department: Mathematics

Section : Secondary

SHEIKH KHALIFA BIN ZAID COLLEGE

Introduction:Practical Geometry will enable the Student to identify and classify geometric figures and apply geometric properties to practical situations, special reasoning and visualization skill will be enhanced by using Geometric Models and a variety of technologies.

• DATE: 27-07-11 Lecture No. 1

• SUBJECT : Mathematics

• TOPIC : Practical Geometry

• CLASS/GROUP: X / IV

• DA, SKBZ College

• LEARNING OBJECTIVES:

• How to Construct “Circum-Circle” of a triangle ABC?

• How to Construct “Escribed-Circle” opposite to the vertex “A” of the triangle ABC?

• How to Construct the “In Circle” (Inscribed Circle) of the ABC?

• How to draw the “Medians” of a ABC?

• How to draw the Altitudes of a triangle?

The teacher asks the following questions.

• How do you measure an angle?

• Do you make an angle of 60° with the help of compass or Protractor?

• How you construct a triangle when its three sides are given?

• How you construct a triangle when the measures of two sides and the included angle is given?

• How you construct a triangle when the measures of two angles and the included side is given?

• What is a line and a line segment.

• What is a triangle?

• Sum of the measures of three angle of a triangle is ……

• Do the sum of the measures of two sides of a triangle is greater than the third side?

• How you bisect an angle?

• How you bisect a line segment?

• What are the vertices of a triangle?

RESOURCE MATERIAL / TRAINING AIDS:

• Board, Lesson Plan, Planner, Chalk, Duster Compass, Protractor, Ruler

By adopting the method of learning by doing and lecture methods of teaching. The teacher will explain how to:

(i) Bisect a line Segment.

(ii) Bisect an angle.

(iii) Make a triangle when.

(a) Three sides are given

(b) Two sides and the included

angle is given.

(c) When two angles and the

included side is given

Students will gain more marks if they learn it with perfection.

Understanding of concept well.

LONG TERM GOALS :

It will be helpful to students in professional studies like mechanical engineering, instrumentation and civil engineering etc

• The Bisectors of Angles of a given triangle.

Given: ABC is a triangle

Required : To draw the bisectors of angles A, B and C

Construction:

1. Draw the ABC

2. With point B as a centre draw an arc of any radius, intersecting the sides BC and BA at point L and M

ANGLES BISECTORS OF  ABC

3. Take point L as a centre and draw an arc of any radius.

4. Now take M as a centre and with the same radius draw the another arc, which cuts the previous arc at point P

5. Join point P to B and produce it. Thus BP is the bisector of <B

6. Similarly, repeat steps (2) to (5) to draw CQ and AR the bisectors of <C and <A respectively.

Hence BP, CQ and AR are the required bisectors of angles of a  arc concurrent.

Note: The angles bisectors intersect at a point I, which is called in centre and the bisectors of angles of a  are concurrent.

• How to draw the right bisectors of sides of a triangle ABC.

Given: A  ABC

Required: To draw right bisectors of the sides AB, BC

and CA

1. Draw the  ABC.

2. To draw right bisector of the side AB, with B as a centre and radius more than half of AB, draw arcs on either sides of AB.

3. Now with A as a centre and with the same radius, draw arcs on either sides of AB cutting previous arcs at P and Q.

4. Join P and Q

Thus PQ is the right bisector of AB,

5. Similarly, repeat steps (2) to (4) and draw ST and LM, the right bisectors of BC and AC respectively. Hence PQ, ST and LM arc the required right bisector of the sides AB, BC and AC respectively of  ABC.

Note: Right bisectors of ABC pass through O. So they arc concurrent.

RIGHT BISECTORS OF  ABC

1. Construction of Circum Circle

Construct the PQR in which mPQ = mQR = 8cm, m<Q =60°. Draw the “Circum Circle” of the  PQR.

Circle:

A set of points of a plane which are equidistant from a given (fixed) point is called a CIRCLE:

The fixed point is called the “Centre” of the Circle.

Circum Circle:

A circle which passes through the three vertices of a triangle is called “Circum Circle” (or circumscribed circle) and the centre is called “Circum Centre” of the circle.

Given:

A  PQR, in which mPQ =mQR = 8cm and m<Q=60°

Required:

Draw the Circum circle of  PQR.

• Draw a line segment QR = 8cm.

• At point Q draw an <RQX = 60°

• From QX, cut off PQ = 8 cm.

• Join P to R. Thus  PQR is constructed.

• Draw the right bisector LL’ of side QR.

• Draw the right bisector MM’ of side PR. Let them intersect each other at point “O”.

• With Centre “O” and radius equal to OR, draw a circle, which passes through the vertices of  PQR.

Hence it is our required circum circle is constructed (constructed)

CIRCUM CIRCLE OF  PQR

2. CONSTRUCTION OF ESCRIBED CIRCLE :

2. Construction of Escribed Circle

Question:

Construct  ABC in which mAB = 7.5cm mBC = 7.5cm, m<B=30° Draw the escribed circle opposite to Vertex B.

Given:

• ABC, in which mAB = mBC = 7.5cm, and m<B=30°

Required:

Draw the escribed circle opposite to Vertex B

• Draw a line segment BC = 7.5cm.

• At point B, draw an < CBX=30°

• From BX, cut off AB = 7.5 cm.

• Join A to C. Thus  ABC is constructed.

• Produce (extend) BC beyond C to BY

• Produce BA to BX. Making two exterior angles <ACY and <CAX.

• Draw the bisector AA’ and CC’ of <CAX and <ACY respectively let them intersect each other at point “O”.

• From point “O” draw OD perpendicular on BY.

• With centre O and radius equal to OD draw a circle. Which touching BA, BY and AC.

This is called the escribed circle opposite to vertex B

ESCRIBED CIRCLE OPPOSITE TO VERTEX “B” OF  ABC

ESCRIBED CIRCLE:

A circle which touches one side of a triangle externally and the other two sides produced internally.

CONSTRUCTION OF INCIRCLE

Question:

Construct the  ABC, in which m<B = 60°, mBC = 7.5cm, mAB = 7cm. Draw the in circle (inscribed circle) of ABC.

Solution

Given:

A  ABC, in which m<B = 60°, mBC = 7.5cm and mAB = 7cm.

Required:

Draw the in circle of ABC.

In-Circle:

The circle which touches the three sides of a triangle is called in-circle or inscribed circle of the triangle and its centre is called “In-Centre”.

• Draw a line segment BC = 7.5cm

• At point “B” draw an <CBX = 60°.

• From BX, cut of AB = 7cm.

• Join A to C. Thus  ABC is constructed.

• Draw the bisector BB’ of >B.

• Draw the bisector CC’ of >C (et them intersect each other at point O.

• From point “O” draw OD perpendicular on BC.

• With Centre “O” and radius equal to OD draw a circle. The circle touches the three sides of ABC

Hence it is our required in-circle is constructed.

Question:

Construct a  ABC, such that mAB = 6.8cm, mBC = 8cm and mAC = 10.0cm. Draw the medians of ABC.

Median of a Triangle:

Line segments joining, the vertices of a triangle to the mid points of the opposite sides are called medians of the triangle.

Given:

A triangle ABC, in which mAB = 6.8cm, mBC = 8cm and mAC = 10.0cm.

Required:

Draw the medians of ABC.

• Draw a line segment BC = 8cm

• From point B draw an arc of radius 6.8cm.

• From point C draw an arc of radius 10cm which intersects the previous arc at point A.

• Join A to B and A to C.

Thus  ABC is constructed.

• Draw the mid point D of Side BC

• Draw the mid point E of Side AC

• Draw the mid point F of Side AB

• Join A to D, B to E and C to F.

Thus AD, BE and CF are our required medians of ABC, which meet in a point O.

It may be noted that medians of every triangle are concurrent (i.e. meet in one point) and their point of concurrency called “centraid”, divides each of them in the ratio of 2:1.

By actual measurement it can be prove that

mAO = mBO = mCO = 2

mOD mOE mOF 1

MEDIANS OF  ABC

CONSTRUCTION OFALTITUDES OF A TRIANGLE

Question:

Take any triangle ABC and draw its altitudes.

Given:

A  ABC

Required:

To draw altitudes of the  ABC

Altitude:

A perpendicular from the vertex of a triangle to the opposite side is called an altitude of the triangle.

1. Draw a  ABC

2. Take a point A as centre and draw an arc of suitable radius, which cuts BC at points D and E.

3. From D as centre, draw an arc of radius more than ½ mDE.

4. Again from point E draw another arc of same radius, cutting first arc at point F.

5. Join the points A and F, such that AF intersects BC at point P. Then AP is the altitude of the  ABC from the vertex A.

6. Similarly, repeat the steps (2) to (5) and draw BQ and CR, the altitudes of  ABC from the vertices B and C respectively.

Hence AP, BQ and CR are the required altitudes of  ABC, which meet in a point O.

The points P, Q and R are called the feet of these altitudes.

Note: It may be remember that the altitudes of a triangle are always concurrent. The point of concurrency is called the “ORTHO CENTRE”

For assessment of the students, the teacher will give them following construction.

• Draw the right bisectors of sides of a triangle ABC, in which mBC = 6.2cm, mAB = 4.5cm and mAC 5cm.

• Construct  ABC in which mAB = 4.5cm, mBC = 5.5cm and m<B = 60°. Draw the circum circle of the  ABC.

• Construct ABC, in which mAB = mBC = 4.5mm and m<B = 30°. Draw the escribed circle opposite to vertex B.

• Draw the bisectors of angles of ABC.

• Construct a ABC in which mAB = 4.5cm mBC = 5.5cm and m<B = 45°. Draw the in-circle of ABC.

• Take any ABC and draw medians of this triangle.

• Take any ABC and draw its altitudes.

• A circle which passes through the vertices of a triangle is called circum-circle.

• A circle which touches the three sides of a triangle is called In-Circle.

• The line segments joining the vertices of a triangle with the mid-points of the opposite sides are called “Medians” of the triangle.

• The point of intersection of three medians is called “Centroid” and medians are concurrent.

• The line segments which are perpendiculars from the vertices to the opposite sides of a triangle are called its “Altitudes”.

• The point of intersection of three altitudes is called “Ortho Centre” and they are concurrent.

• Construction of unique triangle is possible when the sum of measures of two sides is greater than the third side.

Ex No. 9.3 Book 1

P.No 232 Q.No. 1,2,3,5

Ex No. 7.1 Book 2

Q.No. 3

Construct a  ABC in which m AB = 5cm, m<B = 105° and mBC = 4cm.

Draw

i) its circum circle.

ii) Its in-circle

iii) its escribed circle opposite to vertex A.

1) The sum of measures of all the angles of a triangle is __________ degree.

(a) 180 (b) 360 (c) 90 (d) 270

2) Two lines are said to be perpendicular on each other if they form a/an _____ angle.

(a) acute (b) Obtuse (c) Reflex (d) Right

3) The right bisectors of a triangle are ________

(a) Concurrent (b) Not concurrent

(c) Concurrent and passes through the same point (d) None

4) The bisectors of angles of a triangle are and the point of concurrency is called __________

(a) Incentre (b) Ortho Centre

(c) Circum Centre (d) Centroid.

5) A circle which passes through the vertices of a triangle is called __________ .

(a) In-circle (b) Circum-Cricle

(c) Escribed Circle

6) A circle which touches the three sides of a triangle is called __________

(a) Circum circle (b) Escribed circle

(c) In-Circle

7) The point of intersection of three medians is called _____ .

(a) Ortho Centre (b) Centroid

(c) None (d) In-Centre

8) The point of intersection of three altitudes is called ________

(a) Centroid (b) Ortho Centre

(c) In-Centre (d) Curcum Centre

9) The line segments joining the vertices of a triangle with the mid-points of the opposite sides are called __________

(a) Altitudes (b) Medians

(c) Perpendiculars (d) None

10) The line segments which are perpendiculars from the vertices to the opposite sides of a triangle are called ________

(a) Medians (b) Altitudes (c) Bisectors (d) None

Ex. 7.2 Book 2 Page No. 159

TOPICS:

TO BE TAUGHT NEXT:

1) To draw two tangents to a circle from a point outside circle.

2) Direct common tangents to the two given circles.

• To draw Transverse common tangents to the two given circles.

1- (a) 180 2. (d) Right 3. (c) Concurrent and passes through the same point.

4- (a) In-Centre 5. (b) Circum Circle

6. (c) In-Circle 7. (b) Centroid

8. (b) Ortho centre 9. (b) Medians

10. (b) Altitudes

• LESSON PLAN

• DATE: : 28-07-2011

• SUBJECT : Mathematics

• TOPIC : Practical Geometry

• CLASS/GROUP: X / IV

• LECTURE NUMBER: II

• DA, SKBZ College

• SKB - 136

• LEARNING OBJECTIVES:

• To draw a “TANGENT” to a circle from a given point out side the circle.

• To draw the “Direct Common Tangents” to the two given circles.

• To draw the “Transverse Common Tangents” to the two given circles.

• Board

• Lesson Plan

• Planner

• Chalk

• Duster

• Compass

• Protractor

• Ruler

• What is circle?

• What is centre of the circle?

• How you define a radius?

• What is diameter?

• What is chord?

• How you define tangent?

• How you draw a circle. When its radius is given.

• What is circumference of a circle?

• How to draw the right bisector of side (line segment)?

By adopting the method of learning by doing and lecture methods of teaching, The teacher will explain how to:

(i) Draw a circle when its radius is given.

(ii) Bisect a line Segment.

(iii) Draw a tangent to a circle at a point

on it.

(iv) Define a circle and its centre

(b) Diameter of the circle

(c) Chord of the circle

(d) Tangent of the circle

(e) Circumference of the circle

CIRCLE:

A set of points of a plane which are equidistant from a fixed (given) point is called a circle.

The fixed point is called the centre of the circle.

Centre of Circle

The distance of any point of the circle from its centre is called “radius” of the circle. It may be noted.

C A

2cm

Radius is a number not a line segment.

The line segment joining the centre with any point of the circle is called the “Radial Segment”.

In figure we see a circle whose centre is C and radius 2cm and CA is the radial segment.

• CHORD OF THE CIRCLE:

• The line segment whose end points are any two points of the circle is called a Chord of the circle. In fig. CD is the chord.

• DIAMETER:

• A chord which passes through the centre of the circle is called a diameter. In fig PQ is a diameter.

• CIRCUMFERENCE OF A CIRCLE:

• The length of the curve joining all points of a circle is called circumference of the circle.

• Mathematically it is denoted by

• C = 2

• Where  is the radius of the circle and  = 22 or 3.14 7

• : It is the ratio of circumference to its diameter is

called .

C

D

Q

P

If there is only one point Common between a line and a circle, the line is called a tangent to the circle. OR

A line which is perpendicular to the outer end of the radius is called a tangent. OR

A line which touches the circle at only one point is called a tangent. In Fig. PQ is a tangent.

SECANT:

If there are two distinct points common to a line and a circle, the line is called a secant of the circle. In fig. DE is a secant.

SEMI-CIRCLE:

The portion of a circle intercepted by a diameter is called semi-circle.

In the figure diameter AB divides the circle in two equal (congruent) parts. Each is a semi circle.

E

D

P

Q

B

A

1- Question:

Draw a circle of radius 2.5cm Take a point T at a distance of 6cm from the centre. Draw the tangents to the circle passing through T. Measure the lengths of the tangents and verify the answer by mathematical calculation.

Given:

A circle of radius 2.5cm. T is a point out side the circle at a distance of 6cm from the centre of the circle.

Required:

To draw tangents to the given circle from the point-T

Construction:

1. Draw a circle of radius 2.5cm with centre O.

2. Take a point “T” outside the circle such that

OT = 6cm.

3. Draw the mid point M of OT.

4. With centre “M” and radius OM or MT draw a

circle S2 intersecting the given circle “S1” at

points P and Q.

Thus PT and QT are the required tangents.

6. On measuring of PT or QT we get 5.4 cm

VERIFICATION:

In right angled triangle  OPT.

m<OPT = 90° (Angle in the Semi-Circle)

(m OP)2 + (m PT)2 = (mOT)2

(Acc. To Pythagoras theorem)

(2.5)2 + (mPT)2 = (6)2

6.25 + (mPT)2 = 36

(m PT)2 = 36-6.25

(m PT)2 = 29.75

mPT =  29.75

mPT = 5.4 cm

Hence verified

TANGENTS TO A CIRCLE FROM :A POINT OUT SIDE THE CIRCLE

2- Question:

Draw a circle of radius 3cm and 1.5cm. The distance between their centres is 6cm. Draw their direct common tangents. Measure each tangent and verify by mathematical calculation.

Direct Common Tangents to Two Circles:

Definition:

If the points of contact of the common tangent with two circles are on the same side of the line segment joining their centres. Such common tangent is called Direct Common Tangent.

Construction:

1. Draw the line segment AB = 6cm, where A and B

are centres of the two circles.

2. With centre A and radius r1 =3cm draw the first circle S1

3. With centre B and radius 2 = 1.5cm draw the

second circle S2.

4. With centre “A” draw the third circle S3 of radius

3 = 3-1.5 = 1.5cm)

5. Draw the mid-point “M” of AB.

6. With centre “M” and radius equal to AM or MB

draw the fourth circle S4 cutting the third circle

S3 at points C and C’.

7. Join A with C and produce it to meet the first circle S1 at point “D”.

8. Similarly, join A with C’ and produce it to meet the first circle “S1” at point D’.

10. With centre D and radius equal to BC draw an

arc on S2 i.e. E

11. With centre D’ and radius equal to BC draw an

arc on S2 i.e. E’.

12. Draw DE and D’E’.

13. Also Join B to E and E’ .

Thus DE and D’E’ are our required direct common tangents”. On measuring we get

DE = D’E’ = 5.8cm

VERIFICATION:

In  ACB

m<ACB = 90° (Angle in the Semi-circle)

According to Pythagoras theorems

(AB)2 = (AC)2 + (BC)2

(6)2 = (1.5)2 + (DE)2 (since BC = DE)

36 = 2.25 + (DE)2

DE2 = 36 – 2.25

DE2 = 33.75

DE = 33.75

m DE = 5.8cm

Thus verified

Definition:

It the points of contact of the common tangents of two circles lie on the opposite side of the line segment joining the centres of the circles then their tangents are called Transverse Common Tangents.

Question :

Draw the transverse common tangents of the circles with radii 3cm and 2cm, when the distance between their centres is 6cm. Measure each tangent and verify by mathematical calculations.

Given:

Distance between their centres = 6cm.

Required:

Draw Transverse Common Tangents to

these circles.

1. Draw a line segment AB = 6cm the distance between their centres.

2. With centre A, draw the first circle S1 of radius r1 = 3cm.

3. With centre B, draw the second circle S2 of radius r2 = 2cm

4. With centre A, draw the third circle S3 of radius r3 = r1+r2 = 3 + 2 = 5cm.

5. Draw the mid point “M” of AB.

6. With Centre M and radius equal to AM or BM, draw the fourth Circle S4. Which cutting (intersecting) the third circle S3 at points C and C’ respectively.

7. Join A to C and A to C’ which intersect the first circle at D and D’.

8. Join B to C and B to C’

9. With centre D and radius equal to BC draw an arc on the second circle S2 i.e. E

10. With Centre D’ and radius equal to BC draw an arc E’ on the Second Circle S2.

11. Draw DE and D’E’.

12. Also join B to E and B to E’

Thus DE and D’E’ are our required Transverse Common Tangents.

On measuring we get

m DE = mD’E’ = 3.8cm

VERIFICATION:

In  ACB ,

m<ACB = 90° (Angle in the Semi-circle)

According to Pythagoras theorems

(mAB)2 = (AC)2 + (BC)2

(6)2 = ( 5 )2 + (DE)2 ( DE = BC )

36 = 25 + (DE)2

(DE)2 = 36 – 25

(DE)2 = 11

m DE =  11

mDE = 3.3 cm

Thus verified

TESTING / ASSESSMENT / EVALUATION: Common Tangents.

For assessment of the students, the teacher will give them following construction.

1) Draw a circle of radius 2cm. Take a point P at a distance of 5.5cm from the centre of the circle. Draw the tangents to the circle passing through point P. Measure the length of the tangents and verify by mathematical calculation.

2) Draw two circles of radii 2.5cm and 1.2cm, with the distance between their centres equal to 5cm. Draw the Direct Common Tangents to these circles. Also measure and verify by M.C.

3) Draw the Transverse Common tangents of circles with radii 2.5cm and 1.5cm, when the distance between their centres is 5.5cm. Measure each tangent and verify by M.C.

SUMMARY OF THE LESSON: Common Tangents.

1) Set of points of a plane which are equidistant from a fixed point is called a Circle.

2) A chord which passess through the Centre of the Circle is Called a diameter.

3) A line which is perpendicular to the outer end of the radius is called a tangent.

4) Angle inscribed in a Semi-circle is 90°

5) The side opposite to 90° is called hypotenuse.

6) In right angled triangle, according to Pythagoras theorem:

(Hypt.)2 = (Perp.)2 + Base)2

BRIDGE:

The above construction plays a vital role in study of mathematics of H.S.C. i.e. XI & XII. For the study of analytical Geometry. In deed, it is the back bone in study of Architectural and Civil engineering.

ACTIVITIES (M.C.Qs) Common Tangents.

1) The line segment whose end points are any two points of the circle is called a __________ of the circle.

(a) Chord (b) Diameter (c) Radius

(d) Tangent

2) A chord which passes through the centre of the circle is called

(a) Diameter (b) Radius (c) Tangent (d) Secant

3) If a line touches the circle at only one point is called a ________

(a) Tangent (b) Secant (c) Chord (d) Diameter

4) Two circles are said to be congruent if their __________ are equal

(a) Radii (b) Chords (c) Diameters (d) Tangent

5) An angle inscribed in a Semi-Circle is _______ degree.

(a) 90° (b) 180° (c) 60° (d) 360°

6) In right angled triangle, the side opposite to 90 degree is called _______.

(a) Hypotenuse (b) Perpendicular (c) Base (d) Radius

Page -2- Common Tangents.

7) A line which is perpendicular to the outer end of the radius is called a __________

(a) Tangent (b) Secant (c) diameter (d) Chord

8) Two ________ can be drawn to a circle from a point outside the circle.

(a) Tangents (b) Secants (c) Diameters (d) Chords.

9) If the legs of a right angled triangle are 1, 1 then its hypotenuse is _________

(a)  2 (b) 2 (c) ½ (d) 1

10) The line which is at a distance equal to radius of a circle from its centre is ________ to the circle

(c) Diameter (d) Secant