# RD Sharma Solutions Class 8 Practical Geometry Exercise 18.3

## RD Sharma Solutions Class 8 Chapter 18 Exercise 18.3

### RD Sharma Class 8 Solutions Chapter 18 Ex 18.3 PDF Download

#### Exercise 18.3

1. Construct a quadrilateral ABCD in which AB = 3.8 cm, BC = 3.4 cm, CD = 4.5 cm, AD= 5 cm and $\angle B=80°\;$.

Steps of construction:

Step I: Draw AB = 3.8 cm.

Step II: Construct $\angle ABC=80°\;$.

Step III: With B as the center and radius 3.4 cm, cut off BC = 3.4 cm.

Step IV: With C as the center and radius 4.5 cm, draw an arc.

Step V: With A as the center and radius 5.3 cm, draw an arc to intersect the arc drawn in Step IV at D.

Step VI: Join AD, BC and CD to obtain the required quadrilateral.

2. Construct a quadrilateral ABCD, given that AB = 8 cm, BC = 8 cm, CD =10 cm, AD =10 cm and $\angle A=45°\;$.

Steps of construction:

Step I: Draw AB = 8 cm.

Step II: Construct $\angle BAD=45°\;$.

Step III: With A as the centre and radius 10 cm, cut off AD =10 cm.

Step IV: With D as the centre and radius 10 cm, draw an arc.

Step V: With B as the centre and radius 8 cm, draw an arc to intersect the arc drawn in Step IV at C.

Step VI: Join BC and CD to obtain the required quadrilateral.

3. Construct a quadrilateral ABCD in which AB = 7.7 cm, BC = 6.8 cm, CD = 5.1 cm, AD = 3.6 cm and $\angle C=120°\;$.

Steps of construction:

Step I: Draw DC = 5.1 cm.

Step II: Construct $\angle DCB=120°\;$.

Step III: With C as the center and radius 6.8 cm, cut off BC = 6.8 cm.

Step IV: With B as the center and radius 7.7 cm, draw an arc.

Step V: With D as the center and radius 3.6 cm, draw an arc to intersect the arc drawn in Step IV at A.

4. Construct a quadrilateral ABCD in which AB = BC = 3 cm, AD = CD = 5 cm and $\angle B=120°\;$.

Steps of construction:

Step I: Draw AB = 3 cm.

Step II: Construct $\angle ABC=120°\;$.

Step III: With B as the center and radius 3 cm, cut off BC = 3 cm.

Step IV: With C as the center and radius 5 cm, draw an arc.

Step V: With A as the center and radius 5 cm, draw an arc to intersect the arc drawn in Step IV at D.

5. Construct a quadrilateral ABCD in which AB = 2.8 cm, BC = 3.1 cm, CD = 2.6 cm and DA =3.3 cm and $\angle A=60°\;$.

Steps of construction:

Step I: Draw AB = 2.8 cm.

Step II: Construct $\angle BAD=60°\;$.

Step III: With A as the center and radius 3.3 cm, cut off AD = 3.3 cm.

Step IV: With D as the center and radius 2.6 cm, draw an arc.

Step V: With B as the center and radius 3.1 cm, draw an arc to intersect the arc drawn in Step IV at C.

Step VI: Join BC and CD to obtain the required quadrilateral.

6. Construct a quadrilateral ABCD in which AB = BC = 6 cm, AD = DC = 4.5 cm and $\angle B=120°\;$.

Steps of construction:

Step I: Draw AB = 6 cm.

Step II: Construct $\angle ABC=120°\;$..

Step III: With B as the centre and radius 6 cm, cut off BC = 6 cm. Now, we can see that AC is about 10.3 cm which is greater than AD + CD = 4.5 + 4.5 = 9 cm.

We know that sum of the lengths of two sides of the triangle is always greater than the third side but here, the sum of AD and CD is less than AC.

So, construction of the given quadrilateral is not possible.