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Applications of geometrical properties (2.5)

 

2.5  Application of above properties to simple theoretical problems requiring one or more steps of reasoning.” (syllabus)

 

The “above properties” (2.1 to 2.4) are shown below.

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.

 

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