Construction of Quadrilaterals

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 = 72^o at the centre of a circle, at O.

Constructing DAOB, AB = 4 cm, ÐOAB = ÐOBA = 54^o

OA = OB since they are the radii of the circle.

= 54^o

(i) Draw AB = 4 cm.

(ii) At A, make ÐBAX = 54^o 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  = 54^o

Method II:

Constructing each interior angle = 108^o

Exterior angle of a regular pentagon

each interior angle = 180^o - 72^o = 108^o

(i) Draw AB = 4 cm.

(ii) using a protractor, make ÐABX = ÐBAY  = 108^o

(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  = 108^o