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| Construction of Special types of Quadrilaterals(Contd…) |
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| Method 1: |
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| side AB = 3.2 cm |
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| (i) Draw a circle with radius 3.2 cm. |
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| (ii) Take a point A on its circumference. |
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| (iii) With the same radius of 3.2cm, taking A as centre, cut the circle at B. |
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| (iv) From B, with the same radius, cut the circle at C and so on until you obtain 6 points - A, B, C, D, E and F. |
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| (v) Join AB, BC, CD, DE, EF and AF. |
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| (vi) \ABCDEF is the required hexagon. |
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| Method II: |
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| A regular hexagon forms 6 equilateral triangles. |
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A regular hexagon subtends  at the centre |
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  is an equilateral triangle. |
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| Area of hexagon = 6 x Area of DAOB |
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| = 26.60 cm2 |
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| (i) Draw AB = 3.2 cm. |
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| (ii) With centres A and B and radius 3.2 cm, draw two arcs which intersect at O. |
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| (iii) Notice that DAOB is one of the 6 equilateral triangles. |
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| (iv) With centre O, and the same radius of 3.2 cm cut the previous pair of arcs at C and F. |
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| (v) Join AF and BC. |
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| (vi) With centres C and F, and with the same radius of 3.2 cm, cut the previous arcs at D and E. |
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| (vii) Join CD, DE and EF. |
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| (viii) \ABCDEF is the required regular hexagon. |
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| There are two methods of constructing a regular pentagon. |
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| Method I: |
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| AB = 4 cm |
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A pentagon subtends an angle of  = 72o at the centre of a circle, at O. |
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| Constructing DAOB, AB = 4 cm,
ÐOAB = ÐOBA = 54o |
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 OA = OB since they are the radii of the circle. |
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  |
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| = 54o |
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| (i) Draw AB = 4 cm. |
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| (ii) At A, make ÐBAX
= 54o using a protractor. |
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| (iii) Construct the perpendicular bisector of AB which cuts AX at O. |
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| (iv) With O as centre and radius AO, draw a circle. |
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| (v) With centre B and radius = 4 cm, cut the circle at C. |
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| (vi) Again, with centre C and the same radius of 4 cm, cut the circle at D. |
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| (vii) Similarly, with centre D cut the circle at E. |
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| (viii) Join BC, CD, DE and AE. |
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(ix)  ABCDE is the required regular pentagon. |
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| We can mark point O by making
ÐOAB = ÐOBA = 54o |
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| Method II: |
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| Constructing each interior angle = 108o |
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Exterior angle of a regular pentagon  |
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 each interior angle = 180o - 72o = 108o |
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| (i) Draw AB = 4 cm. |
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| (ii) using a protractor, make
ÐABX = ÐBAY =
108o |
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| (iii) Cut BX and AY at C and E respectively, so that BC = AE = 4 cm. |
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| (iv) With centres C and E and radius 4 cm, draw two arcs, which intersect at D. |
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| (v) Join CD and DE. |
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| \ ABCDE is the required regular pentagon. |
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| Instead of step (iv), we can draw
ÐAED = ÐBCD =
108o |
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