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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