NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2.

- Practical Geometry Class 8 Ex 4.1
- Practical Geometry Class 8 Ex 4.3
- Practical Geometry Class 8 Ex 4.4
- Practical Geometry Class 8 Ex 4.5

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 8 |

Subject |
Maths |

Chapter |
Chapter 4 |

Chapter Name |
Practical Geometry |

Exercise |
Ex 4.2 |

Number of Questions Solved |
1 |

Category |
NCERT Solutions |

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

**Ex 4.2 Class 8 Maths Question 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

**(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

**Solution.**

**(i)** Let us draw a rough sketch of the required quadrilateral and write down the dimensions. Clearly, the two easily constructible triangles are UT and LIF.

Steps of Construction :

- Draw LI = 4 cm.
- With L as centre and radius 2.5 cm, draw an arc.
- With I as centre and radius 4 cm draw another arc to cut the previous arc at T.
- Join TL and 77.
- With L as centre and radius 4.5 cm, draw an arc.
- With I as centre and radius 3 cm, draw another arc to cut the previously drawn arc at F.
- Join FI, FL and TF.

Then, LIFT is the required quadrilateral.

**(ii)** Let us draw a rough sketch of the required quadrilateral and write down its dimensions. Clearly, the two easily constructible triangles are LOD and LGD.

Steps of Construction :

- Draw LD = 5 cm.
- With L as centre and radius 7.5 cm, draw an arc.
- With D as centre and radius 10 cm, draw another arc to cut the previous arc at O.
- Join OL and OD.
- With L as centre and radius 6 cm, draw an arc.
- With D as centre and radius 6 cm, draw an arc to cut the previously drawn arc at B.
- Join GD, GL and OG.

Then, GOLD is the required quadrilateral.

**(iii)** We know that the diagonals of a rhombus bisect each other at right angles. So, we proceed according to the following steps.

**Steps of Construction :
**

- Draw BN = 5.6 cm.
- Draw the right bisector XY of BN, meeting BN at O.
- From O set off OE = \(\frac { 1 }{ 2 } \) (6.5) cm = 3.25 cm along OY and OD = 3.25 cm along OX.
- Join BE, EN, ND and DB.

Then, BEND is the required rhombus.

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