Construct the following quadrilaterals.

(i) Quadrilateral LIFT


LI = 4 cm


IF = 3 cm


TL = 2.5 cm


LF = 4.5 cm


IT = 4 cm


(ii) Quadrilateral GOLD


OL = 7.5 cm


GL = 6 cm


GD = 6 cm


LD = 5 cm


OD = 10 cm


(iii) Rhombus BEND


BN = 5.6 cm


DE = 6.5 cm

(i) A rough sketch of this quadrilateral:


(a) Δ ITL can be constructed as follows:



(b) Vertex F is 4.5 cm away from vertex L and 3 cm away from vertex I.


Hence, taking L and I as centers, draw arcs of 4.5 cm radius and 3 cm radius respectively, which will intersect each other at point F



(c) Join F to T and F to I



LIFT is the required quadrilateral


(ii) A rough sketch of this quadrilateral:



(a) Δ GDL can be constructed as follows:



(b) Vertex O is 10 cm away from vertex D and 7.5 cm away from vertex L.


Hence, taking D and L as centers, draw arcs of 10 cm radius and 7.5 cm radius respectively These will intersect each other at point O



(c) Join O to G and L



GOLD is the required quadrilateral


(iii) We know that the diagonals of a rhombus always bisect each other at 90°


So in this case let us assume these are intersecting each other at point O


Therefore,


EO = OD = 3.25 cm


A rough sketch of this rhombus:



(a) Draw a line segment BN of 5.6 cm and its perpendicular bisector. Let it will be intersecting the line segment BN at point O



(b) Taking O as centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at point D and E



(c) Join points D and E to points B and N



BEND is the required quadrilateral


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