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Plane Geometry – Geometrical Properties (2.1 – 2.4)

 

2.1       Preliminaries on diagrams, notation, symbols and conventions.

2.2       Definitions of special plane figures.

2.3       Properties of angles at a point and by angles formed by transversals to parallel lines. Tests for parallel lines.

Angle sums of triangles, quadrilaterals and general polygons.

Exterior angle properties.

Congruence of triangles. Tests for congruence.

Properties of special triangles and quadrilaterals. Tests for special quadrilaterals.

Properties of transversals to parallel lines.

Similarity of triangles. Tests for similarity.

Pythagoras’ theorem and its converse.

Area formulae.

2.4  Application of above properties to the solution of numerical exercises requiring one or more steps of reasoning.” (syllabus)

 

For a more detailed description of the requirements for this topic, see the mathematics syllabus on the Board of Studies website click here

 

 

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GEOMETRY I

 

Q.1.  Find the value of the pronumerals in the following diagrams. Give a reason for your answer in each case.

Q.2. Find the values of the pronumerals in the following diagram.

                       

Q.3.  Find the value of the pronumerals in the following diagrams. Give a reason for your answer in each case.

        

 

Q.4.     Show that AB is parallel to CD.

                       

 

ANSWERS

Q.1.     a = 74 (alternate angles are equal); b = 60 (co-interior angles are supplementary); c = 54 (alternate angles are equal)

Q.2.     a = 70, b = 70, c = 110, d = 110, e = 70, f = 110.

Q.3.     (i) a = 60 (exterior angle of triangle = sum of interior opposite angles

130 = 70 + a    a = 60)

            b = 50 (BCD is a straight angle = 180o  b = 180 – 130 = 50)

 (ii)       a = 50 (angle sum of D = 180o) b = 50  (BAC = 90o)

            c = 40 (angle sum of DBAC = 180o)     e = 140 (DCE straight line)

Q.4.     DCB = 120o (sum of angles in DBCD = 180o)

            ABC = 120o (given)

            Since DCB = ABC and are alternate then AB || CD

 

GEOMETRY II

 

Q.1.     Find the values of the pronumerals a & b and hence show AB||ED.

Q.2.     Find the values of the pronumerals a, b & c and hence show that DAFE is isosceles.

Q.3.     Triangles ABC and DBC are isosceles. Find the values of the pronumerals a, b & c, giving reasons for your answers.

ANSWERS

 

Q.1.     a = 100 (sum of angles in DDEC = 180o)

            b = 50 (sum of angles in DABC = 180o)

            Since ABC = EDC (each 50o) and these are complementary angles then ED || AB

Q.2.     ACB = 80o (BCD is a straight angle = 180o)

            a = 50o (angle sum of DABC = 180o)

            AEF = ABC = 50o (corresponding angles) b = 50

            c = 80 (angle sum of DAEF = 180o)

            Since a = b = 50o then DAEF is isosceles with EF = AF (base angles equal)

Q.3.     BCD = 30o (base angles of isosceles triangle are equal)

            a = 120 (angle sum of DDBC = 180o)

            b = 360 – 120 = 240 (angle sum at a point = 360o)

            BCA = 30 + 20 = 50o (base angles of isosceles triangle are equal)

            c = 80 (angle sum of DABC = 180o)

 

SIMILAR FIGURES

  1. A tree casts a shadow of 6.4 metres when a 3.0 metre shadow stick casts a shadow of 1.6 metres. How tall is the tree?

 

  1. Later in the day, the stick in Q.1. cast a shadow of 2.1 metres. What would be the length of the shadow cast by the tree at this time?

 

 

  1. A student who is 160 cm tall casts a shadow of 40 cm at the same time as the school building casts a shadow of 4.6 metres. How tall is the school building?

 

  1. Explain, in words, how you could determine the height of a telegraph pole.

 

 

  1. Josh and Ryan wanted to find the diameter of the Sun. They knew that the Sun is 150 million km from Earth. Josh poked a pinhole in a piece of cardboard while Ryan drew two lines 0.5 cm apart on a page of his exercise book. Ryan held a metre rule above his exercise book with the zero mark just next to the lines he drew on his book. Josh moved the cardboard with a pinhole up and down next to the ruler and focused the Sun’s image on Ryan’s book. When the image was focused on the lines and was exactly 0.5 cm in diameter Josh measured the distance between the cardboard and the book. It was 49.8 cm. Draw a diagram showing the two similar triangles and calculate the diameter of the Sun.

 

  1. A transparency with a square of side 6.0 cm was placed on an overhead projector. A square of side 48 cm was produced on the screen. What was the ratio of the area of the image to the area of the original square?

 

Answers:

1. 12m             2. 8.4m            3. 18.4m

 

4. Measure the length of the shadow cast by the telegraph pole. Hold a 1 metre ruler vertically on level ground. Use a second ruler to measure the length of the shadow cast by the 1 metre ruler.

Height of telegraph pole =

5.  

Dia. = 1.506 million km = 1.506 x 106km

 

6.         Ratio = 64:1

 

 

 

Geometry III

 

1.            ABC is an equilateral triangle.  BC is produced to D so that BC = CD.  CE is drawn so that it bisects ACD and is equal to CD.  AE is joined.

(i)                  Draw a diagram of the above figure.

(ii)                Show that ABCE is a rhombus.

(iii)               Show that ACDE is a rhombus.

 

2.            ABCD is a straight line with AB = BC = CD. E, F & G are 3 points on the same side of AD such that AEB, BFC & CGD are equilateral triangles.

(i)                  Draw the above figure.

(ii)                Show that EFG is a straight line.

(iii)               Show that AEGC is a parallelogram.

        

3.            ABCD is a parallelogram. E & F are two points on the diagonal DB such that

         DE = EF = FB.

(i)                  Draw the above figure.

(ii)        Show that ΔAED ≡ ΔCFB

 

4.      AB & CD are equal in length and AB || CD.  AD & BC intersect at E.

(i)         Draw the above figure.

(ii)                Show that ΔAEB ≡ ΔCED

 

5.      ABC is an isosceles triangle with AB = AC.  The side BC is produced to D so that AC = CD. ABC = q.                                                                                    (i) Draw the above figure.

      (ii) Show that ADC =                                                                                          

6.      ABCD is a rhombus. E is the mid-point of AB and F is the mid-point of DC.                                (i) Draw the above figure.

      (ii) Show that DADE ≡ DBCF

 

7.            ABC is an isosceles triangle with AB = BC. D is a point inside the triangle and DB = DC.                                                                                                    (i) Draw the above figure

(ii)    Show that ABD = ACD                                                                                

8.   ABCD is a square. E is a point inside the square such that ED = EC.                                               (i) Draw the above figure.

      (ii) Show that DADE ≡ DBCE

9.   ABCD is a quadrilateral with AB = CD. Diagonals AC and DB are drawn to intersect at E. CAB = BDC

            (i) Draw the above figure.

            (ii) Show that DABE  DDCE

      (ii) Hence show that DBAC ≡ DBDC

10.    ABCD is a square. E is a point on the side BC and F is a point on the side DC such that CE = FC.

            (i) Draw the above figure.

            (ii) Show that DABE ≡ DADF                                                                                                 

 

Geometry 3.

1.(i)     

 

(ii)        ABC is an equilateral triangle AC =AB = BC

            ACE is an equilateral triangle AC = CE = AE

            AB = BC = CE = AE

            ABCE is a rhombus (quadrilateral with all sides equal.)

 

   (iii)    ECD is an equilateral triangle EC =CD = DE

            ACE is an equilateral triangle EC = AC = AE

            AC = CD = DE = AE

            ACDE is a rhombus (quadrilateral with all sides equal.)

 

2. (i)    

            (ii)        EAB & FBC are both equilateral triangles (given)

             ACB = ECD = 60o

            EBF = 60o (ABC is a straight angle)

            And since EB = EF (given AB = BC in equilateral triangles)

            Then BEF = BFE

            Also BEF + BFE = 180o- 60o = 120o (angle sum of triangle)

            So EFB = half of 120o = 60o

            Similarly it can be shown that GFC = 60o

            EFB + BFC + GFC = 60o + 60o + 60o = 180o

            EFG is a straight line.

 

    (iii)   Since all triangles are equilateral with some common sides then all sides of the triangles are equal

            EF + FG = AB + BC

            Also EA = GC

            AEGC is a parallelogram (opposite sides equal)

           

3. (i)    

(ii)        AB = BC (opposite sides of parallelogram)

            ADE = DBC (alternate, BC || AD)

            DE = FB (given)

            DAED  DBFC (SAS)

 

4. (i)    

(ii)        BAE = CDE (alternate, AB || CD)

            ABE = DCE (alternate, AB || CD)

            AB = CD (given)

            DAEB  DCED (AAS)

 

5. (i)

(ii)        ABC = ACB = q (base angles of isosceles triangle)

            ACD = (180 - q)  (BCD is a straight angle)

            CAD = ADC (base angles of isosceles triangle given that AC = CD)

            ACD +CAD +ADC = 180 (sum of angles in triangle)

            Then CAD +ADC = q

            i.e. CAD = ADC =

6. (i)    

(ii)        AD = BC (opposite sides of rhombus)

            AE = FC (E & F are mid-points of equal sides)

            DAE = BCF (diagonally opposite angles of rhombus)

            DADE  DBCF (SAS)

7. (i)    

(ii)        ABD = ACB (base angles of isosceles triangle)

            DBC = DCB (base angles of isosceles triangle)

            ABD - DBC = ACB - DCB

            i.e. ABD = ACD

 

8. (i)    

(ii)        EDC = ECD (base angles of isosceles triangle)

            ADC = BCE = 90o (angles in a square)

            ADE = BCE (complements of equal angles)

            Also AD = BC (sides of square)

            And ED = EC (given)

            DADE  DBCE (SAS)

 

9. (i)    

(ii)        DEC = AEB (vertically opposite)

            CDB = CAB (given)

            DC = AB (given)

            DABE  DDCE (SAS)

 

    (iii)   AE = ED (DABE  DDCE)

            EC = EB (DABE  DDCE)

            AE + EC = DE + EB

            i.e. AC = DB

            Also CDB = CAB (given)

            And DC = AB (given)

            DBAC  DBDC  (SAS)

 

 

10. (i)  

(ii)        Since BC = CD (sides of square)

            And CE = CF (given)

            Then BC – CE = CD – CF

            i.e. BE = DF

            Also AB = AD (sides of square)

            And ABE = ADF = 90o (angles of square)

            DABE  DADF (SAS)