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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