Polygons are everywhere! A polygon is any 2-dimensional shape formed with straight lines. Triangles, quadrilaterals, pentagons, and hexagons are all examples of polygons. The name tells you how many sides the shape has. For example, a triangle has three sides, and a quadrilateral has four sides. So, any shape that can be drawn by connecting three straight lines is called a triangle, and any shape that can be drawn by connecting four straight lines is called a quadrilateral.

A polygon is a plane figure formed by joining three or more straight sides. A polygon is said to be regular if all its sides are equal and its angles are equal.

**Types of Polygon**

A **pentagon** is a polygon with five sides.

A** hexagon** is a polygon with six sides.

A **heptagon** has seven sides.

An **octagon** has eight sides.

A **decagon** has ten sides.

**Construction of a Polygon**

**(A) To construct a regular a Regular Hexagon given its side**

**(I) Using 60 ^{0} set square**

- Draw a horizontal line and mark off AB equal to the side of the hexagon.

- Through A, draw a line at 60
^{0 }and mark off AC equal to AB.

iii. Through B, draw a line at 60^{0} parallel to BD and mark off BD equal to AB.

- Through C, draw a line of 60
^{0 }parallel to BD and mark off CE equal to AB. - Through D, draw a line of 60
^{0 }parallel to AC and mark off DF equal to AB. - Join EF to complete the hexagon

**(II) Using a pair of Compass**

This method is best remembered as the constant Radius Rule.

- Draw a circle whose radius is equal to the side of the hexagon. Draw the horizontal diameter AB.
- With centre A and the same radius, cut the circle above AB at E and below AB at F.

iii. With centre B and the same radius, cut the circle above AB at E and below AB at F.

- Join AD, DF, FB, BE, EC, and CA to obtain a hexagon.

Note: This procedure is required to draw a regular hexagon given the distance across corners. The diameter of the circle is equal to the distance across corners.

**(B) To construct a regular hexagon given the distance across flats**

- Draw a circle whose diameter is equal to the distance across flats. Draw the vertical diameter AB.
- Draw diameter CD and EF at 30
^{0}.

iii. Through A and B, draw horizontal tangents.

- Through C, D, E, F, in turn, draw tangents at 60
^{0}. The figure that is formed by the intersection of the tangents is the required hexagon.

**Hexagon**

Note: This is the procedure when it is required to describe a regular hexagon about a given circle.

**(C) To construct a regular octagon given its side**

- Through C and D, draw vertical lines and mark off CE and DF equal to AB.
- Through E and F, draw lines at 45
^{0}and mark off EG and FH equal to AB.

iii. Join GH to complete the octagon.

- Through C and D, draw vertical lines and mark off CE and DF equal to AB.
- Through E and F, draw lines at 45
^{0}and mark off EG and FH equal to AB. - Join GH to complete the octagon.

**(D) To construct a regular octagon given the distance across flats**

- Draw a circle whose diameter is equal to the distance across flats. Draw horizontal diameter AB and vertical diameter CD.
- Draw diameters EF and GH at 45
^{0}.

iii. Draw vertical tangents through A and B and horizontal tangents through C and D.

- Through E, F, G H in turn draw tangents at 45
^{0}. The figure formed by the intersection of the tangents is the required octagon.

Note: This is the procedure when it is required to describe a regular about a given circle.

**(E) General methods for constructing a regular polygon on a given base**

**(a) The ‘External – 360 ^{0}/N Rule**’

- Obtain the external angle of the required polygon by dividing 360
^{0}by the number of sides (N) of the polygon i.e. external angle = 360^{0}/N. - Draw a horizontal line and mark off AB equal to the given base.

iii. Through A, draw a line at 360^{0}/N and mark off a length equal to AB. Also N at B, draw a line at 360^{0}/N and mark off a length to AB.

- Continue the process until you have obtained the polygon N side where N = 5, 6, 7, 8, 9, 10, ….

Suppose that N = 5, then external angle = 360^{0}/N = 72^{0}. The pentagon will be obtained by drawing at 72^{0}(b).https://googleads.g.doubleclick.net/pagead/ads?guci=2.2.0.0.2.2.0.0&client=ca-pub-1661929042807246&output=html&h=280&slotname=9757662299&adk=416771370&adf=2768648106&pi=t.ma~as.9757662299&w=700&fwrn=4&fwrnh=100&lmt=1604309876&rafmt=1&psa=1&format=700×280&url=http%3A%2F%2Fstoplearn.com%2Fcourses%2Fsecondary-school%2Fjss2-second-term-basic-technology-junior-secondary-school%2Flessons%2Fplane-figure%2F&flash=0&fwr=0&fwrattr=true&rpe=1&resp_fmts=3&wgl=1&adsid=ChAIgPn-_AUQkJWp-eO5jtJNEkgAMWjS4Gv1WpgKnpn8nZ948lg6520CC2BFMxTmQ9r7DwfLvhB5w_EeNvMBSwDz0TzzHkj7SNUELj32g2J1dDRG9dIVe1EpeaY&tt_state=W3siaXNzdWVyT3JpZ2luIjoiaHR0cHM6Ly9hZHNlcnZpY2UuZ29vZ2xlLmNvbSIsInN0YXRlIjowfSx7Imlzc3Vlck9yaWdpbiI6Imh0dHBzOi8vYXR0ZXN0YXRpb24uYW5kcm9pZC5jb20iLCJzdGF0ZSI6MH1d&dt=1604309765645&bpp=8&bdt=141&idt=73&shv=r20201029&cbv=r20190131&ptt=9&saldr=aa&abxe=1&cookie=ID%3D3b5c0a4b352ccd7b-2255262b41a60032%3AT%3D1602230088%3ART%3D1602230088%3AS%3DALNI_MaWUG7yiaaHthyGgrREyU5I0uWdRw&prev_fmts=0x0%2C1200x90_0ads_al&nras=1&correlator=1648033421824&frm=20&pv=1&ga_vid=1467971727.1604309766&ga_sid=1604309766&ga_hid=1853458312&ga_fc=0&iag=0&icsg=599886135210&dssz=59&mdo=0&mso=0&u_tz=60&u_his=21&u_java=0&u_h=768&u_w=1366&u_ah=728&u_aw=1366&u_cd=24&u_nplug=3&u_nmime=4&adx=609&ady=661&biw=1518&bih=667&scr_x=0&scr_y=0&eid=21065724%2C44730557&oid=3&psts=AGkb-H-ksd2NH0A2cEBKqa9nXZhLjieCOHIPiGHBVWHaxhfq6Fm_H6B1og&pvsid=1798998084752492&pem=804&ref=http%3A%2F%2Fstoplearn.com%2Fcourses%2Fsecondary-school%2Fjss2-second-term-basic-technology-junior-secondary-school%2Flessons%2Fquadrilaterals%2F&rx=0&eae=0&fc=896&brdim=0%2C0%2C0%2C0%2C1366%2C0%2C1366%2C728%2C1517%2C666&vis=1&rsz=%7C%7CaoeE%7C&abl=CA&pfx=0&fu=8320&bc=31&jar=2020-10-31-22&ifi=1&uci=a!1&fsb=1&xpc=0d3YTGLrOF&p=https%3A//stoplearn.com&dtd=M

**(b) The ‘Two-Triangle Rule**’

- Draw a horizontal line and mark off AB equal to the given base.
- Bisect AB and produce its bisector as long as it is convenient.

iii. On AB as base, draw an isosceles triangle with base angle 45^{0} and an equilateral triangle so that the apexes of the two triangles lie on the bisector of AB. Denote the apex of the isosceles triangle as d, and that of the equilateral triangle as f.

- Bisect fd to obtain point e.
- Along the bisector of AB, from the point f, step off length de (or ef) to obtain points g, h, I, j, etc. The points d, e, f, g, I, j are the centres of the circumscribing circles for a square, regular pentagon, hexagon, heptagon, octagon, nonagon and decagon respectively.
- suppose you want to draw a polygon of 9 sides (nonagon). With centre I and radius I A (or iB) draw a circle. Take length AB and step it off on the circle to obtain the points, C, D, E, F, G, H, I.

Join the points to obtain the required regular nonagon.

(Observed that d = 4; e = 5; f= 6; g = 7; h = 8; i = 9; j = 10.)

**(F) The General Method for Describing a Regular about a given Circle**

The method is best remembered as the ‘Centre – 360^{0}/N Rule’.

- Obtain the angle included by any two normals at the centre of the given circle by dividing 360
^{0}/N by the number of sides N of the required described polygon, i.e. angle at centre = 360^{0}/N. - Draw the given circle with centre O and draw a vertical radius OA.

iii. Use a protractor to set out angles of 360^{0}/N and draw radii of OB, OC, OD etc., until you have got N radii.

- Through the points A, B, C, D, etc., draw tangents to obtain the required polygon. Suppose it is required to draw an octagon.

360^{0}/N = 360^{0}/8 = 45^{0}.

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