# Exercise 4.2

Question 1

LI = 4 cm

IF = 3 cm

TL = 2.5 cm

LF = 4.5 cm

IT = 4 cm

Sol :

A rough sketch of this quadrilateral can be drawn as follows.

(Step-1) Δ ITL can be constructed by using the given measurements as follows.

(Step-2) Vertex F is 4.5 cm away from vertex L and 3 cm away from vertex I. Therefore, while taking L and I as centres, draw arcs of 4.5 cm radius and 3 cm radius respectively, which will be intersecting each other at point F.

(Step-3) Join F to T and F to I.

OL = 7.5 cm

GL = 6 cm

GD = 6 cm

LD = 5 cm

OD = 10 cm

Sol :

A rough sketch of this quadrilateral can be drawn as follows.

(Step-1) Δ GDL can be constructed by using the given measurements as follows.

(Step-2) Vertex O is 10 cm away from vertex D and 7.5 cm away from vertex L. Therefore, while taking D and L as centres, draw arcs of 10 cm radius and 7.5 cm radius respectively. These will intersect each other at point O.

(Step-3) Join O to G and L.

(iii) Rhombus BEND

BN = 5.6 cm

DE = 6.5 cm

Sol :

We know that the diagonals of a rhombus always bisect each other at 90º. Let us assume that these are intersecting each other at point O in this rhombus.

Hence, EO = OD = 3.25 cm

A rough sketch of this rhombus can be drawn as follows.

(Step-1) Draw a line segment BN of 5.6 cm and also draw its perpendicular bisector. Let it intersect the line segment BN at point O.

(Step-2) Taking O as centre, draw arcs of 3.25 cm radius to intersect the perpendicular bisector at point D and E.

(Step-3) Join points D and E to points B and N.

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