Class 8 | NCERT Solution Maths Chapter 4 | Practical Geometry | Exercise 4.2

Exercise 4.2 Page: 62

1. Construct the following quadrilaterals.

(i) Quadrilateral LIFT LI = 4 cm

IF = 3 cm TL = 2.5 cm LF = 4.5 cm

IT = 4 cm

Solution:

A rough sketch of the quadrilateral LIFT can be drawn as follows.

(1) ∆ ITL can be constructed by using the given measurements as follows.

(2) Vertex F is 4.5 cm away from vertex L and 3 cm away from vertex I. ∴, 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.

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

LIFT is the required quadrilateral.

(ii) Quadrilateral GOLD OL = 7.5 cm

GL = 6 cm GD = 6 cm LD = 5 cm OD = 10 cm

Solution:

The rough sketch of the quadrilateral GOLD can be drawn as follows.

(1) ∆ GDL can be constructed by using the given measurements as follows.

(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.

(3) Join O to G and L.

GOLD is the required quadrilateral.

(iii) Rhombus BEND

BN = 5.6 cm

OD = 6.5 cm

Solution:

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

The rough sketch of the rhombus BEND can be drawn as follows.

(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.

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

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

BEND is the required quadrilateral.