Rectilinear Figures


   
 
Construction of Special types of Quadrilaterals(Contd…)
Construction of a Regular Hexagon (Given the Side)
 
Method 1:
 
 
side AB = 3.2 cm
 
 
(i) Draw a circle with radius 3.2 cm.
 
(ii) Take a point A on its circumference.
 
(iii) With the same radius of 3.2cm, taking A as centre, cut the circle at B.
 
(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.
 
(v) Join AB, BC, CD, DE, EF and AF.
 
(vi) \ABCDEF is the required hexagon.
 
 
Method II:
 
A regular hexagon forms 6 equilateral triangles.
 
 
 
A regular hexagon subtends at the centre
 
is an equilateral triangle.
 
Area of hexagon = 6 x Area of DAOB
 
 
 
= 26.60 cm2
 
 
(i) Draw AB = 3.2 cm.
 
(ii) With centres A and B and radius 3.2 cm, draw two arcs which intersect at O.
 
(iii) Notice that DAOB is one of the 6 equilateral triangles.
 
(iv) With centre O, and the same radius of 3.2 cm cut the previous pair of arcs at C and F.
 
(v) Join AF and BC.
 
(vi) With centres C and F, and with the same radius of 3.2 cm, cut the previous arcs at D and E.
 
(vii) Join CD, DE and EF.
 
(viii) \ABCDEF is the required regular hexagon.
 
 
Construction of a Regular Pentagon
 
There are two methods of constructing a regular pentagon.
 
Method I:
 
 
AB = 4 cm
 
A pentagon subtends an angle of = 72o at the centre of a circle, at O.
 
Constructing DAOB, AB = 4 cm, ÐOAB = ÐOBA = 54o
 
 
OA = OB since they are the radii of the circle.
 
 
                                 = 54o
 
 
(i) Draw AB = 4 cm.
 
(ii) At A, make ÐBAX = 54o using a protractor.
 
(iii) Construct the perpendicular bisector of AB which cuts AX at O.
 
(iv) With O as centre and radius AO, draw a circle.
 
(v) With centre B and radius = 4 cm, cut the circle at C.
 
(vi) Again, with centre C and the same radius of 4 cm, cut the circle at D.
 
(vii) Similarly, with centre D cut the circle at E.
 
(viii) Join BC, CD, DE and AE.
 
(ix) ABCDE is the required regular pentagon.
 
 
 
We can mark point O by making ÐOAB = ÐOBA  = 54o
 
Method II:
 
Constructing each interior angle = 108o
 
 
Exterior angle of a regular pentagon
 
each interior angle = 180o - 72o = 108o
 
 
(i) Draw AB = 4 cm.
 
(ii) using a protractor, make ÐABX = ÐBAY  = 108o
 
(iii) Cut BX and AY at C and E respectively, so that BC = AE = 4 cm.
 
(iv) With centres C and E and radius 4 cm, draw two arcs, which intersect at D.
 
(v) Join CD and DE.
 
 
\ ABCDE is the required regular pentagon.
 
 
Instead of step (iv), we can draw ÐAED = ÐBCD  = 108o
 
 
     
   
Get FREE Live Tutoring
Get FREE Live Tutoring
(No credit card required)

Customer Care

Click to get customer service, technical support and subscription help.

Customer Care Chat


Refer-A-Friend

Get One Month Free!
When you refer a friend