# NCERT Solutions for Class 8 Maths Chapter 4- Practical Geometry Exercise 4.2

The NCERT Solutions for Class 8 Maths enhance topics with frequent, focused, engaging math challenges and activities that strengthen math concepts. Each question of the exercise has been carefully solved for the students to understand, keeping the examination point of view in mind.

Class 8 Maths Chapter 4-Practical Geometry Exercise 4.2 questions and answers helps the students to understand the construction of a quadrilateral when two diagonals and three sides are given. These NCERT Solutions are prepared by BYJUâ€™S subject experts using a step-by-step approach.

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### Access Other Exercise Solutions of Class 8 Maths Chapter 4- Practical Geometry

Exercise 4.1 Solutions 1 Question

Exercise 4.3 Solutions 1 Question

Exercise 4.4 Solutions 1 Question

Exercise 4.5 Solutions 4 Questions

### Access Answers to NCERT Class 8 Maths Chapter 4- Practical Geometry Exercise 4.2 Page Number 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) âˆ† 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. âˆ´, 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.

#### (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) âˆ† 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.

(iii) Rhombus BEND

BN = 5.6 cm

DE = 6.5 cm

Solution:

We know that the diagonals of a rhombus always bisect each other at 90Âº.

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.