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NCERT Solutions Class 8 Mathematics Solutions for Exercise 4.2 in Chapter 4 - Practical Geometry

Question 1 Exercise 4.2

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

Answer:

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

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

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

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

(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.

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

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

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

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

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

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

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

(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.

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

(3) Join O to G and L.

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

(iii) We know that the diagonals of a rhombus always bisect each other at 90^{\circ}.

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.

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

(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.

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

(2) Taking O as centre, draw arcs of 3.25 cm radius to intersect the perpendicular

bisector at point D and E.

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

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

NCERT Solutions - Mathematics , Class 8 chapter Practical Geometry Question 1 Solution image

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