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Polygons are closed figures with three or more straight sides. If all the sides are the same length, and all the internal angles are equal, we call them regul a r polygons. Triangles are three-sided polygons, and an equil a ter a l triangle is a regul a r three-sided polygon. A square is a regular four-sided polygon. Pent agons have five sides, hexagons have six sides and heptagons have seven. Make a list of as many of these special names as you can find.

Here are several closed plane figures. Decorate them and write the name of each polygon on the shape.

B. Investig a tion

Choose four polygons from the group above, all regular, but with four different numbers of sides. Now measure the sizes of the internal angles of each. Try to find out whether it is possible to make a formula to tell you how large the angles are, and what they add up to.

The following table will be helpful. As you can see, there are infinitely many polygons.

No. of sides a = internal angle size b = 360 – a c = b – 180 Total of a Total of c
Three a = c =
Four a = c =
Five a = c =
Six a = c =
Seven a = c =
Twelve 12× a = 12× c =
  • The characteristics in the table above are needed when you have to decide how to tile a floor (say) with regular polygons so that they don’t overlap and don’t leave gaps. Some of these polygons will work alone, and some can or must be combined.
  • Design and draw a repeating tiling pattern of your own, using only regular polygons and colour it so that the pattern shows clearly.

C. Three-dimension a l closed figures

  • If these shapes have sides made up of polygons, then we call them polyhedr a. A regul a r polyhedron has faces that are congruent regular polygons, with internal angles the same shape and size.
  • In contrast to the polygons, there are only five regular polyhedra. They have been known since the time of Plato and the Greek mathematicians; this is why they are known as the five Platonic Solids.

D. Project

Research the five Platonic Solids, finding their names and properties, and other interesting deductions and facts about them. Make an attractive poster or models of these solids showing the facts associated with each. Below are pictures of the five solids.

Assessment

LO 3
Space and Shape (Geometry)The learner will be able to describe and represent characteristics and relationships between two-dimensional shapes and three-dimensional objects in a variety of orientations and positions.
We know this when the learner :
3.1 recognises, visualises and names geometric figures and solids in natural and cultural forms and geometric settings, including:3.1.1 regular and irregular polygons and polyhedra;3.1.2 spheres;3.1.3 cylinders;3.2 in contexts that include those that may be used to build awareness of social, cultural and environmental issues, describes the interrelationships of the properties of geometric figures and solids with justification, including:3.2.1 congruence and straight line geometry;3.3 uses geometry of straight lines and triangles to solve problems and to justify relationships in geometric figures;3.4 draws and/or constructs geometric figures and makes models of solids in order to investigate and compare their properties and model situations in the environment;
3.5 uses transformations, congruence and similarity to investigate, describe and justify (alone and/or as a member of a group or team) properties of geometric figures and solids, including tests for similarity and congruence of triangles.

Memorandum

Discussion

  • This guide includes two pages of figures for constructing simple right prisms. Photocopy enough for the learners to make at least two of the figures. It would be best if the copies could be made on very light card (or heavy paper). If they are asked to colour some of the parts (e.g. the base and top) it might make it easier to explain some of the more difficult formulae.
  • The two formulae for right prisms are, in general:
  • Total Surface Area = double the base area + height of prism × perimeter of base
  • Volume = base area × height of prism
  • Ensure that learners are clear on the units (squared or cubed) appropriate to each formula.
  • Another difficulty that learners might encounter is that the word height is used in calculating the area of triangles as well as being one of the dimensions of right prisms. A useful trick is to use h for the triangle case and H for the prism case.
  • Breaking down the steps required for the calculations is a useful method for learners who get confused by the components in the formula. Of course, very competent learners will substitute values straight into the formula. This is an effective system, and should be encouraged where appropriate.

Solutions – exercise:

Rectangular prism: TBO = 412 cm 2 Vol = 480 cm 3

Triangular prism: TBO = 307,71 cm 2 Vol = 360 cm 3

Cylinder: TBO = 402,12 cm 2 Vol = 603,19 cm 3

Granny’s Jam Pot: Vol = 8 595,40 cm 2

11 Square–based jars: Vol = 8 096 cm 2

11 Rectangle–based jars: Vol =8 633,63 cm 2

So, granny must use the rectangular–based jars if she wants to fit all the jam in!

3 = triangle; 4 = tetragon; 5 = pentagon; 6 = hexagon; 7 = heptagon; 8 = octagon; * = not polygon

No of sides a = internal angle size b = 360° – a c = b – 180° Total of a Total of c
Three 60° 300° 120° a = 180° c = 360°
Four 90° 270° 90° a = 360° c = 360°
Five 108° 252° 72° a = 540° c = 360°
Six 120° 240° 60° a = 720° c = 360°
Seven 308,57° 51,43° –128,57° a = 2160° c = –360°
Twelve 330° 30° –150° 12× a = 3960° 12× c = –360°

TEST 1

1. Explain how you would recognise a right prism.

2. Explain how you could find the base of a right prism.

3. Calculate the total surface area and the volume of each of the following three prisms. Give your answers accurate to two decimal places.

TEST 1 – Memorandum

1. Essential points in the explanation: three-dimensional; top and base congruent plane shapes; side(s) at right angles to base.

2. Any reasonable explanation, e.g. if the chosen side is placed at the bottom, the description of a right prism fits what you see.

3. Rectangular right prism: TBO = 1 939,68 cm 2 Volume = 5 769,72 cm 3

Triangular right prism: TBO = 1 507,74 mm 2 Volume = 2 312 mm 3

Cylinder: TBO = 8 022,37 m 2 Volume = 41 593,67 m 3

Questions & Answers

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Assimilatory nitrate reduction is a process that occurs in some microorganisms, such as bacteria and archaea, in which nitrate (NO3-) is reduced to nitrite (NO2-), and then further reduced to ammonia (NH3).
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This process is called assimilatory nitrate reduction because the nitrogen that is produced is incorporated in the cells of microorganisms where it can be used in the synthesis of amino acids and other nitrogen products
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Prevent foreign microbes to the host
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they provide healthier benefits to their hosts
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They are friends to host only when Host immune system is strong and become enemies when the host immune system is weakened . very bad relationship!
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cell is the smallest unit of life
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cell is the structural and functional unit of life
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Source:  OpenStax, Mathematics grade 9. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col11056/1.1
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