## RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2

These Solutions are part of RD Sharma Class 8 Solutions. Here we have given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2

**Other Exercises**

- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.1
- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2
- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.3
- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.4
- RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.5

**Question 1.**

**Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.0 cm, AD = 2.3 cm, AC = 4.5 cm and BD = 3.8 cm.**

**Solution:**

**Steps of construction :**

(i) Draw a line segment AB = 3.8 cm.

(ii) With centre A and radius 2.3 cm and with centre B and radius 3.8 cm draw arcs intersecting each other at D.

(iii) Join AD and BD.

(iv) Again with centre A and radius 4.5 cm and with centre B and radius 3 cm, draw arcs intersecting each other at C.

(v) Join AC and BC and also CD.

Then ABCD is the required quadrilateral.

**Question 2.**

**Construct a quadrilateral ABCD in which BC = 7.5 cm, AC = AD = 6 cm, CD = 5 cm and BD = 10 cm.**

**Solution:**

**Steps of construction :**

(i) Draw a line segment CD = 5 cm.

(ii) With centre C and D and radius 6 cm, draw line segments intersecting each other at A.

(iii) Join AC and AD.

(iv) Again with centre C and radius 7.5 cm and with centre D and radius 10 cm, draw arcs intersecting each other at B.

(v) Join CB, CA, DA, DB and AB.

Then ABCD is the required quadrilateral.

**Question 3.**

**Construct a quadrilateral ABCD, when AB = 3 cm, CD = 3 cm, DA = 7.5 cm, AC = 8 cm and BD = 4 cm.**

**Solution:**

**Steps of construction :**

This quadrilateral is not possible as

BD = 4 cm, AB = 3 cm and AD = 7.5 cm

The sum of any two sides of a triangle is greater than the third side.

But BD + AD = 4 + 3 = 7 cm

BD + AD < AD

**Question 4.**

**Construct a quadrilateral ABCD given AD = 3.5 cm, BC = 2.5 cm, CD = 4.1 cm, AC = 7.3 cm and BD = 3.2 cm.**

**Solution:**

**Steps of construction :**

(i) Draw a line segment CD = 4.1 cm.

(ii) With centre C and radius 7.3 cm and with centre D and radius 3.5 cm, draw arcs intersecting each other at A.

(iii) Join AC and AD.

(iv) Again with centre C and radius 2.5 cm and with centre D and radius 3.2 cm, draw arcs intersecting each other at B.

(v) Join CB’, and DB’ and join AB’.

Then ABCD is the required quadrilateral.

**Question 5.**

**Construct a quadrilateral ABCD given AD = 5 cm, AB = 5.5 cm, BC = 2.5 cm, AC = 7.1 cm and BD = 8 cm.**

**Solution:**

**Steps of construction:**

(i) Draw a line segment AB = 5 cm.

(ii) With centre A and radius 7.1 cm and with centre B and radius 2.5 cm, draw arcs which intersect each other at C.

(iii) Join AC and BC.

(iv) Again with centre A and radius 5 cm and with centre B and radius 8 cm, draw arcs which intersect each other at D.

(v) Join AD and BD and CD.

Then ABCD is the required quadrilateral.

**Question 6.**

**Construct a quadrilateral ABCD in which BC = 4 cm, CA = 5.6 cm, AD = 4.5 cm, CD = 5 cm and BD = 6.5 cm.**

**Solution:**

**Steps of construction:**

(i) Draw a line segment CD = 5 cm.

(ii) With centre C and radius 5.6 cm and with centre D and radius 4.5 cm, draw arcs which intersect each other at A.

(iii) Join AC and AD.

(iv) Again with centre C and radius 4 cm and with centre D and radius 6.5 cm, draw arcs which intersect each other at B.

(v) Join BC and BD and AB.

Then ABCD is the required quadrilateral.

Hope given RD Sharma Class 8 Solutions Chapter 18 Practical Geometry Ex 18.2 are helpful to complete your math homework.

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