Lesson Notes By Weeks and Term - Senior Secondary School 1

CONSTRUCTIONS

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SUBJECT: MATHEMATICS

CLASS:  SS 1

DATE:

TERM: 3rd TERM

REFERENCE BOOKS

  • New General Mathematics SSS 1 by M.F. Macrae et al 
  • Essential Mathematics SS 1

 

 
WEEK FOUR

TOPIC: CONSTRUCTIONS

 

  • Construction of quadrilateral polygon i.e. four sided figure with given certain conditions parallelogram 
  • Construction of equilateral triangle 
  • Locus of moving points including equidistance from two lines of two points and constant distance from the point. 

 

 

 

  • Construction of Quadrilaterals

 

 

Examples

  1. Construct a quadrilateral ABCD in which AB is parallel to DC /AB/= 4cm, /BC /= 5cm and /DC/= 7cm and o..Measure  the diagonal BD.
  2. Use your ruler and compasses to construct the parallelogram PQRS in which /QR/ = 5cm, /RS /=11cm and < QRS = 135o.
  3. Measure  the length of the shorter diagonal of PQRS. 

 

Solutions

First make a sketch of the quadrilateral to be constructed as shown in the figure below:












Steps of the required construction are stated as follows:

  1. Draw DNC = 7cm with DN = 3cm and NC = 4cm
  2. Construct CDM = 105o

iii. With N as centre, radius 5cm draw an arc to cut DM at A

  1. With A as centre and radius of 4cm draw an arc.

With C as centre and a radius of 5cm draw a second arc to cut the first arc at B

  1. Join A to B and  C to B to complete the quadrilateral ABCD.

By measurement , /BD/= 4.5cm 















2)First make a sketch of the parallelogram PQRS











The step of the construction are stated as follows:

  1. Draw line QR = 5cm
  2. Construct R = 135o
  3. With R as centre and radius 11 cm draw an arc to cut the angle 135o line at S.
  4. With S as centre and radius 5cm, draw an arc
  5. With Q as centre and radius 11cm, draw a second arc to cut the arc of step iv. This is point P
  6. Draw lines to join S to P and P to Q
  7. Draw dotted line through diagonal RP and measure it.





















By measurement the length of the shorter diagonal PR is 8.7cm 



EVALUATION

  1. Construct quadrilateral ABCD such that /AB/ = 5cm, /BD/= /DC/ =8cm,o and o.
  2. Measure the diagonal /AC/.

 

 

  • Construction of Equilateral Triangle

 

An equilateral triangle is a triangle in which all the sides are of equal length and each of its angle is 60o.





Examples

  1. Construct an equilateral triangle  XYZ such that /XY/= 5CM
  2. (a) Construct an equilateral triangle ABC such that /AB/= 7cm

(b) Construct the bisectors of A, B and C

(c) What do you observe?





Solutions

Sketch:









2) Sketch: The required construction is 


















  1. The bisectors of each angle meet each other at a point inside the equilateral  triangle.

 

 

  • Construction of Loci of Moving Points

 

  1. Locus of points at a given distance from a fixed point.

 

In the figure below, O is a fixed point, Pi, P2 are at a constant distance x cm from O .  The locus of the points is a circle of radius x cm.(see the figure below).










ii).  Locus of point at a given distance from a straight line









In the figure above AB is a straight line which continues indefinitely  in both directions. Points Pi, P2, P3, P4 are each a distance x cm from AB.  In two dimensions, the locus of  the points  consist of two straight lines parallel to AB, each at a distance x cm from AB.

Note that this locus consist of two separate lines.

 

iii.)  Locus of points equidistant from two given points.















In the figure  above, x and y are two fixed points . Points Pi, P2, P3 are such that /PiX/ = /PiY/, /P2X/= /P2Y/and /P3Y. /.  P1, P2, P3, lie on the perpendicular bisector of XY.  The  locus of the points is the perpendicular  bisector of XY (shown in the figure above).



  1. iv)  Locus of Points Equidistant from two straight lines.















In the figure above, AB and CD are straight lines which intersect at O. P1 is equidistant from AB and CD . Similarly, P2 is equidistant from the two lines. P1 and P2 lie on the bisector of the acute angle between the two lines.














In the figure above, P3 is equidistant from AB and CD.P3 lies on the bisector of the obtuse angle between the two lines.

Thus, the complete locus of points which are equidistant from two straight liens is the pair of bisectors of the angles between the lines.( see the figure below).

Note that the two parts of the locus intersect at right angles.















Example

Using ruler and compasses only

a, Construct         ABC such that  /AB/ = 6cm, /AC/ = 8.5cm and  BAC = 120o

 

  1. Construct the locus l1 of points equidistant from A and B,

 

  1. Construct the locus l2 of points equidistant from AB and AC.

 

d.Find the points of intersection P1 and P2, of l1 and l2 and measure /P1 P2/

 

Solution 























  1. Note the construction of BAC = 120o.

 

  1. l1 is the perpendicular  bisector of AB

 

  1. l2 is in two parts.  AP1 is the bisector of BAC. AP2 is perpendicular to AP1, Note that points on AP2 are equidistant from AB and CA produced.
  2. By measurement /P1P2/ =  6.8cm

 

EVALUATION 

  1. Construct an equilateral triangle ABC such that /AB/= 8cm
  2. Construct the midpoints of AB, BC, and CA
  3. What do you observe?

 

READING ASSIGNMENT 

NGM SS BK 1 pages 176-186 Ex 16e No.6 page 186.

 

GENERAL EVALUATION

  1. Construct a     XYZ in  which    /YZ/ = 8.2cm, XYZ = 45o and XZY  = 75o.

measure !XY!.

  1. Using ruler and compasses only, construct:
  1. The locus of a point equidistant from Y and Z.
  2. A point Q on this locus, equidistant from YX and YZ.

 

WEEKEND ASSIGNMENT 

  1. A circle centre O, radius 5cm is drawn on a sheet of paper. A point P moves on the paper so that it is always 2cm from the circle .  The locus of O     A. a circle, centre O, radius 3cm     B. two circles,centre O radii 3cm and 7cm     C. a circle, centre O, radius 6cm      D. two circles,centreO,radii 4cm and 6cm      E. a circle, centre O, radius 3.5cm.
  2. XYZ is a straight line such that /XY/ =/YZ/= 3cm .A point P moves in the plane of XYZ so that /PY/ < /XY/, which of the following describes the locus of P?     A. line through X perpendicular to XZ     B. line through Y perpendicular to XZ     C. line through Z perpendicular to XZ      D. circular disc, centre X,radius 3cm      E. circular disc, centre 4, radius 3cm.
  3. Describe the  locus of a point which moves so that it  is always 5cm from a fixed point O in a plane.    A. rectangle which measures 10cm by 5cm     B. square of side length 5cm      C. a parallelogram whose diagonals are 10cm and 5cm      C. a circle of radius 5cm, centre O       E. a circle of radius 10cm, centre O.
  4. Describe the locus of a point which moves along a level floor so that it is 2m from a wall of a room.A. One line, parallel to and 2m from the wall.      B. Two lines, one each side of, parallel to and 2m from the wall     C. A circle of radius 2m      D. A semi-circle of radius ½ m      E. Two perpendicular lines, each of length 2m
  5. Describe the  locus of a point which moves so that it is 3cm from a fixed line AB in a plane.     A. 2 lines parallel to AB and 6cm apart, joined by semi-circular ends.     B. 2 lines parallel to AB and 8cm apart; joined by semi-circular ends       C. 2 lines perpendicular to AB       D. A circle of radius 6cm      E. circle of radius 3cm.

 

THEORY

  1. construct a trapezium ABCD in which AB is parallel to DC, AB =4cm  BC = 8cm, CD = 11cm, DA = 6cm.  (hint: in a rough figure, divide the trapezium into parallelogram AB X D and triangle BCX. (First construct triangle BCX )
  2. Using ruler and compasses only, construct       
  1. ABC such that   /AC/ = 8.5cm and ACB = 135o.
  2. Using any geometrical instruments, find a point P within ABC which is at a distance 2.8cm from AC and 6cm from B. Measure the length of AP.














  WEEK 5                                                                                    DATE................................................

TOPIC: Deductive proof

Sum of angles in a triangle

The sum of the angles of a triangle is 180.

The sum of the angles of a triangle is 180.





Given any triangle ABC

To prove: A+B+C=180

Construction:Produce BC to a point X.Draw CP parallel to BA.

Proof:With the lettering of the figure above

         a1=a2         (alternate angles)

         b1=b2     (corresponding angles)

c+a1+b1  =  180

       C+a2+b2  =  180

ABC  +  A  +  B =  180

A  +  B  +  C  =  180




Relationship to angles on a straight line

The sum of angles on a straight line is 180o.

The sum of angles on a straight line is 180o.




                                    P + q + r =  180o

Angles on a parallel line cut by a transversal line

The figure below is parallel lines cut by a transversal line indicating angles a – h

Corresponding Angles

From the figure above, the following angles are corresponding:

         a = g    ;   b = h   ;   c = e   ;   d = f

Alternate Angles

From the figure above, the following angles are alternate

         a = d  ;   b = c 




Vertically Opposite Angles

From the figure above, the following angles are vertically opposite

            a = f ;    b = e ;   c = h ;   d = g




Example

Isosceles triangles ABC and ABD are drawn on opposite sides of a common base AB. If ABC= 70 and ADB = 118, calculate ACB and CBD.





Solution







In triangle ABC,

    ABC = 70          (given)

    BAC = 70          (base angles of isos. Triangle)

Therefore, ACB = 180 – 70 – 70    (angle sum of triangle)

                            = 40

In triangle ABD,

        ADB = 118     (given)

Therefore, ABD + BAD = 180 – 118   (angle sum of triangle)

                                        = 62

Therefore, 2 X ABD = 62   (base angles of isos. Triangle)

                           ABD = 31

   CBD = CBA + ABD = 70 + 31 = 101

    ACB = 40 and CBD = 101

 

Parallelogram

A parallelogram is a quadrilateral which has both pairs of opposite sides parallel.

  1.                                                                                       b)      




Rhombus, rectangle and square are special examples of parallelogram. A rhombus is a parallelogram with sides of equal length.

Properties of Parallelogram

  1. i) The opposite sides are parallel.
  2. ii) The opposite sides are equal.

iii) The opposite angles are equal.

  1. iv)  The diagonals bisect one another.

Properties Of Rhombus

  1. i) All four sides are equal.
  2. ii) The opposite sides are parallel.

iii) The opposite angles are equal.

  1. iv)  The diagonals bisect one another at right angles.
  2. v) The diagonals bisect the angles.

NB: In a rectangle, all of the properties of a parallelogram are found and all four angles are right angles. In a square, all of the properties of a rhombus are found and all four angles are right angles.

Intercept





In the figure above, the lines AB and CD cut the transversal PQ into three parts. The part of the transversal cut off between the lines is called an intercept. In the figure above, the line segment XY is the intercept

Intercept Theorem

If three or more parallel lines cut off equal intercepts on a transversal, then they cut off equal intercepts on any other transversal.





Given: Three parallel lines cutting a fourth line at A, B, C so that /AB/=/BC/ and cutting another line at X, Y, Z respectively.

To prove:/XY/ = /YZ/.

Construction: Draw XP and YQ parallel to ABC to cut BY and CZ at P and Q respectively.

Proof:

AXPB is a parallelogram   (opp. Sides //)

             XP = AB                   (opp side equal)                     

Similarly  /YQ/ = /BC/        (in //gm YQCB)

             /XP/ = /YQ/             (given AB = BC )

In triangles XPY, YQZ

             /XP/ =/YQ/               (Proved)

                X1 = x2           (corr. angles)

                Y1 = y2            (corr. angles)

Therefore, triangle XPY =  triangle YQZ  (AAS)

                                    /XY/ = /YZ/  

 

EVALUATION

Find the length k, m, n in the figures below






Congruent Triangles

Two figures or triangles are congruent if they have exactly the same shape and size.The following are conditions for congruency:

i)Two sides and the included angle of one are respectively equal to two sides and the included angle of the other.(SAS) e.g in the figures below, triangle ABC is congruent to PQR  



ii)Two angles and a side of one are respectively equal to two angles and the corresponding side of the other.(ASA or AAS) e.g. the figures below are congruent





iii)The three sides of one are respectively equal to the three sides of the other.(SSS)






iv)They are right-angled, and have hypotenuse and another side of one respectively equal to the hypotenuse and another side of the other.(RHS)










EVALUATION

State whether the triangles are congruent, not congruent or not necessarily congruent. If congruent state condition of congruency







READING ASSIGNMENT

Essential Mathematics for Senior Secondary Schools 1 page 323

 

GENERAL EVALUATION 

  1. In the figure below, <ABP = <110o and o. Calculate BPC









  1. In triangle ABC, o and o. BC is produced to X. the bisectors of
  2. Find the lettered lengths in cm. 












WEEKEND ASSIGNMENT

In each pairs of triangles a), b), c), state the condition of congruency 






  1. State the condition of congruency for the pairs of triangle in a)ASA       b)SAS    c)SSS     d)not congruent
  2. State the condition of congruency for the pairs in b)a)SSS       b)SAS       c)AAS       d)not congruent
  3. State the condition of congruency for the pairs in c)a)SSS    b)SAS    c)RHS  d)not congruent

Use this figure to answer questions 4 and 5





  1. Calculate the angle marked ua)28     b)38     c)48    d)56
  2. Calculate the angle marked va)28       b)56    c)152  d)162

                                     THEORY

  1. Given the data of figure below, prove that triangle PQR is isosceles.





  1. (a) In figure below, a) what is the ratio /AD/ ÷ /DB/ ?

(b) If /DB/ = 5cm, what is /AB/?











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